Organic Chemistry

[graham]

It's too hard to memorize this...especially in come to type reaction...I saw once in the web people posting have the full aliphatic compound mind-mapping regarding to the reaction, reagent used and i'd downloaded it...
But i can't find those in aromatic compound...as some of the reaction and reagent are different from aliphatic
Anyone can help??
Thanks...:))

[Skyevenz]

u cant simply read organic chem. U gt to do more execise n u will noe how's d question ask n wad tactic u can use to tackle the problem.

[music_freak28]

How about constructing your own mind map? It is not that hard if you keep updating it as you learn more..although it can be a bit messy :)) Plus, you'll be able to understand it better because it's your way of presenting it and by writing them down, you can memorize better.

Doing exercise would be the other option.:)

[fridge]

hey, here some questions about organic chemistry i hope can find some explanation to me. =p

1. Hot acidified potassium dichromate(VI) can be used to differentiate between
1 C6HOH and C6H11OH
2 C6H5OH and C6H5CH2CH2OH
3 C2H5OH and CH2=CH2

2. when a compound, W, is reacted with aqueous sodium hydroxide, the product formed doenst decolourise alkaline potasium manganate (VII). Which of the following compounds is W?
A. C6H5CH2Cl
B. (C2H5)3CBr
C. CH3CH2Br
D. CH3CHBrCH2CH3

thanks.

[imforumer]

For question no. 2, it's about oxidation of alcohol. Only tertiary alcohol will not be oxidised.
So the answer must be B.

Reaction process:
(C2H5)3CBr +NaOH(aq) -> (C2H5)3COH (tertiary alcohol)

[Age_2_Perfection]

for Q1
1. Hot acidified potassium dichromate(VI) can be used to differentiate between
1 C6HOH and C6H11OH
2 C6H5OH and C6H5CH2CH2OH
3 C2H5OH and CH2=CH2

i'm not sure what's the "1" option but i assume its C6H5OH (phenol) and C6H11OH (cyclohexanol)

If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions.

Hot acidified potassium dichromate(VI) can only oxidise 1* and 2* alcohols. (primary and secondary).
Therefore, C6H11OH, C6H5CH2CH2OH & C2H5OH can be oxidised to cause a colour change.

BUT at the same time, potassium dichromate(VI) is also a test for the double bond in alkenes, leading to the same colour change. Therefore you cannot differentiate between C2H5OH and CH2=CH2.

Phenols are rather easily oxidized despite the absence of a hydrogen atom on the hydroxyl bearing carbon. Among the colored products from the oxidation of phenol by chromic acid is the dicarbonyl compound para-benzoquinone (also known as 1,4-benzoquinone or simply quinone) - http://www.cem.msu.edu/~reusch/VirtualText/alcohol2.htm

**However this is not within our STPM syllabus. So in this case we assume phenol cannot be oxidised**

So the answer is no.1 and no.2. No.3 is wrong.

[DonFrank]

Hi , since this is a thread about organic chemistry, I would like to share the way I study organic chemistry.

I've tried making mindmaps before but frankly, they don't work effectively because everytime when exam is close, you have to cram them over and over again.

So the way I use now is to read only without making any notes. I try to understand the intuition behind all kinds of reaction. Why this reaction occurs, why they use pumice to eliminate H and Cl in 1,2-dichloroethane and in a quite similar case they use sodium hydroxide in ethanol to eliminate H and Cl in chloroethane.

When your brain knows why these occur or that occur, it is easy now. You don't need to spend a lot of time making a beautiful mindmap already. Everything is inside the brain. You can hence, solve any questions.

This is how I learn organic chemistry, hope it helps you if you are not into making mindmaps. :)

[Athersin]

i have a question here:

what is the type of hybridisation of NO2- and NO and NO2?

Thanks!

[DonFrank]

In nitrite ions, the nitrogen atom forms double bond with an oxygen atom and a single bond with another oxygen atom which carries a negative charge. In fact, nitrite ions exists as a resonance hybrid. Nitrogen uses sp hybridisation since it forms a single bond and a double bond.

In nitrogen monoxide, the nitrogen forms a double bond with oxygen atom by sharing its two electron with oxygen atom. It violates the octet rule since it has only 11 valence electrons. The hybridisation used by nitrogen is s hybridisation. Head-on overlapping occurs between the s hybrid orbital of nitrogen and p orbital of oxygen atom. The unhybridised p-orbital of nitrogen overlaps sideway with the p orbital of oxygen atom to form pi bond.

In nitrogen dioxide, it is a resonance hybrid too but the oxygen atom does not carry any negative charge. Sp hybridisation too.

[chongkeat]

So the way I use now is to read only without making any notes. I try to understand the intuition behind all kinds of reaction. Why this reaction occurs, why they use pumice to eliminate H and Cl in 1,2-dichloroethane and in a quite similar case they use sodium hydroxide in ethanol to eliminate H and Cl in chloroethane.
Indeed, why pumice? I've been curious about the mechanism, but just not curious enough.
UPDATE:
In fact, what the heck is pumice anyway. There's just "volcanic rock" in the Ace Ahead book...
Wikipediaaaa! (http://en.wikipedia.org/wiki/Pumice)

what is the type of hybridisation of NO2- and NO and NO2?

Hey, that question came out for my trials! Got 0 for that one, naturally. If only I had seen this question sooner...



Mindmaps are kinda useful, but only as a test of your knowledge. My technique involves reading through the entire chapter, and trying to construct a sort of, erm, chapter taxonomy, that lists out the reactions and their mechanisms, with some extra info scribbled around the sides.

Then I make sure i got everything down correctly, and throw everything away. Then, revise the previous chapter the same way.

Lather, rinse, repeat.

[Athersin]

In nitrite ions, the nitrogen atom forms double bond with an oxygen atom and a single bond with another oxygen atom which carries a negative charge. In fact, nitrite ions exists as a resonance hybrid. Nitrogen uses sp hybridisation since it forms a single bond and a double bond.

In nitrogen monoxide, the nitrogen forms a double bond with oxygen atom by sharing its two electron with oxygen atom. It violates the octet rule since it has only 11 valence electrons. The hybridisation used by nitrogen is s hybridisation. Head-on overlapping occurs between the s hybrid orbital of nitrogen and p orbital of oxygen atom. The unhybridised p-orbital of nitrogen overlaps sideway with the p orbital of oxygen atom to form pi bond.

In nitrogen dioxide, it is a resonance hybrid too but the oxygen atom does not carry any negative charge. Sp hybridisation too.

Is the hybridisation only been counted as s or p for those bonding pairs rite? Nonbonding pairs are not counted as a hybridised orbitals just because they don't have any orbitals which are side overlap ( pi Bond) or head-on overlap( Sigma bond)? Am i right?

I don't really get why NO is s hybridasation? it is so weird to see this type of unusual hybridisation as i never encountered before. haha..
if this is the case, N=O, so the nitrogen atom should have -ve charge right due to the nonbonding electron pairs, rite?

for NO2,the structure is O=N=O rite? so N aton should carry +ve charge? Should we show the orbital with the +ve charge?




Hey, that question came out for my trials! Got 0 for that one, naturally. If only I had seen this question sooner...


haha, so coincident:)), this is the question i encountered when i am doing revision for inorganic chemistry book, the thick international book. forget its names, the ans stated is sp3 for all these 3 molecules and ions, weird la, even the world international book can have such a great mistake. :wink

[DonFrank]

Is the hybridisation only been counted as s or p for those bonding pairs rite? Nonbonding pairs are not counted as a hybridised orbitals just because they don't have any orbitals which are side overlap ( pi Bond) or head-on overlap( Sigma bond)? Am i right?

I don't really get why NO is s hybridasation? it is so weird to see this type of unusual hybridisation as i never encountered before. haha..
if this is the case, N=O, so the nitrogen atom should have -ve charge right due to the nonbonding electron pairs, rite?

for NO2,the structure is O=N=O rite? so N aton should carry +ve charge? Should we show the orbital with the +ve charge?



haha, so coincident:)), this is the question i encountered when i am doing revision for inorganic chemistry book, the thick international book. forget its names, the ans stated is sp3 for all these 3 molecules and ions, weird la, even the world international book can have such a great mistake. :wink

Correcion, hybridisation occurs only during formation of two or more covalent bonds.

Read the post below.

[chongkeat]

We consider double and triple bonds as a single bonding pair.
Since in NO, there is a double bond, we say it has only got a single bonding pair, therefore it must have undergone s hybridisation. [ If 2 bonding pair, then sp hybridisation and if 3, then sp2 hybridisation and if 4, then sp3 and so on sp3d1, sp3d2 etc]Single bonding pair? You mean as in a single bond?

*Non bonding pair does no undergo hybridisation. They can be lone pair and usually they will be used in the formation of pi bond. Sigma bond only involves electrons hybridised orbitals.Are non bonding pairs equivalent to lone pairs? Or are lone pairs just a subset of non bonding pairs?
Or rather, what are non bonding pairs anyway?

For NO2-... it looks like this too.. (-)O-N=O but the former Oxygen atom carries a negative charge.
Just wondering:
How does the extra electron fit in the general structure of the molecule?
Does it affect the resonance in any way?

UPDATE:
Got this from the web.
[image removed due to size issue. Do reformat the image so that people can view it without scrolling. Original image link (http://z.about.com/d/chemistry/1/0/T/W/lewisnitrite.jpg)] *seehua~

Hmm, does this mean that the extra electron is sucked into the resonance, as in it travelling back and forth?

[Athersin]

nitrogen atom in NO does not carry any negative charge. It is simply electron deficient. It has only got 7 valence electron after forming a double covalent bond with oxygen atom. This usually happens when the atom has odd number of valence electrons (nitrogen has 5 valence electrons).


If N forms double bond with O, N=O, then N suppose to have -ve charge, isn't it? N is neutral when it has 3 bonding pairs, while if it got only 2 bonding pairs as in N=O, the N suppose to be negatively charged, rite?
This is what i learnt from the lecturers, would u mind explain further? i dun really get it.


For NO2, it is O-N=O but in fact it is a resonance hybrid.

Resonance hybrid is always a stable compound, but still the former O is -ve charge..How is it to become a neutral NO2??

*For NO2, the nitrogen atom does not lose any electron so it is not positively charged. It is just that the bonds are polar which mean oxygen attract the electrons more strongly than nitrogen.

okay, in this case, i agree that the N atom is neutrally charged since it has 3 bonding pairs. But isn't the charge for the former O atom is -ve charge?


For NO2-... it looks like this too.. (-)O-N=O but the former Oxygen atom carries a negative charge.


yea, it should be like that as in the NO2 , isnt't it?

Really confused now.
Hope to get some enlightenment.:)

[yeng]

Single bonding pair? You mean as in a single bond?


Are non bonding pairs equivalent to lone pairs? Or are lone pairs just a subset of non bonding pairs?
Or rather, what are non bonding pairs anyway?

single bonding pair = 2 atoms bonded together, no matter how many bonds between them...
non bonding e- pairs = non bonding pairs.

Just wondering:
How does the extra electron fit in the general structure of the molecule?
Does it affect the resonance in any way?

UPDATE:
Got this from the web.
[image removed, see original post]

Hmm, does this mean that the extra electron is sucked into the resonance, as in it travelling back and forth?


the charge is evenly distribute between the whole molecule, however lewis structure cannot display the correct structure. sometimes u can see dotted line display between
the bend.

the molecule of NO2 is neutral, therefore it is assumed everything is not charged. FOrmal charge(charge of the atoms in molecules) is ignored unless it is in degree level.

[DonFrank]

single bonding pair = 2 atoms bonded together, no matter how many bonds between them...
non bonding e- pairs = non bonding pairs.




the charge is evenly distribute between the whole molecule, however lewis structure cannot display the correct structure. sometimes u can see dotted line display between
the bend.

the molecule of NO2 is neutral, therefore it is assumed everything is not charged. FOrmal charge(charge of the atoms in molecules) is ignored unless it is in degree level.

I think Yeng has clarified all your doubts. Thanks Yeng.

[chongkeat]

Okay. First of all, I sincerely apologize for putting up that pic.

And one last question:
We consider double and triple bonds as a single bonding pair.
Since in NO, there is a double bond, we say it has only got a single bonding pair, therefore it must have undergone s hybridisation. [ If 2 bonding pair, then sp hybridisation and if 3, then sp2 hybridisation and if 4, then sp3 and so on sp3d1, sp3d2 etc]
Does this apply in general, or only in some specific cases, as in NO?

Take for example, CO2. It is double bonded to two Oxygen atoms, so 2 bonding pairs. Does that mean it undergoes sp hybridisation?

[DonFrank]

Okay. First of all, I sincerely apologize for putting up that pic.

And one last question:

Does this apply in general, or only in some specific cases, as in NO?

Take for example, CO2. It is double bonded to two Oxygen atoms, so 2 bonding pairs. Does that mean it undergoes sp hybridisation?

Exactly. It is applied to all chemical compounds.
Another example is N2...There is a triple bond but we still consider it as a single bonding pair.

[DonFrank]

Okay. First of all, I sincerely apologize for putting up that pic.

And one last question:

Does this apply in general, or only in some specific cases, as in NO?

Take for example, CO2. It is double bonded to two Oxygen atoms, so 2 bonding pairs. Does that mean it undergoes sp hybridisation?

It applies to every molecules when we want to predict their molecular geometry.

i have a question here:

what is the type of hybridisation of NO2- and NO and NO2?

Thanks!

Ignore what I've said before since my answers are inaccurat and fallacios.
I've spent some time to relearn all the bonding and now I've been very clear about them. Hopefully my explanation below will help you.

Nitrite ion
In NO2(-), the nitrogen forms a single covalent bond with an oxygen atom which carries a negative charge and a double covalent bond with a neutral oxygen atom. [There is a lone pair in nitrogen atom]

Nitrogen atom has 5 valence electrons. Its valence electronic configuration is 2s2 2px1 2py1 2pz1. When it forms nitrite ions, it undergoes sp2 hybridisation, ie, 2s2 2px1 2py1 undergo hybridsation to form three hybridised sp2 orbitals, leaving one unhybridised 2pz1 orbitals.

formation of single covalent bond in nitrite ion
The sp2 hybrid orbital of nitrogen with an unpaired electron will head-on overlap with 2px1 orbital of negatively charged oxygen atom to form sigma bond.

formation of double covalent bond
The sp2 hybrd orbital with an unpaired electron of nitrogen will head-on overlap with 2px1 orbital of neutral oxygen atom to form sigma bond while the unhybridsed 2pz orbital of nitrogen will sideway overlap with another 2py1 orbital of the neutral oxygen atom to form pi bond. This gives rise to the double covalent bond in nitrite ion.

lone pair
The lone pair is in fact the sp2 hybrid orbital with paired electrons.

Hence, we say that nitrogen atom in NO2(-) has undergone sp2 hybridisation and it is predicted to have a bent structure.

Nitrogen dioxide
Nitrogen dioxide exists as a radical in nature. So we can expect that nitrogen in nitrogen dioxide to has an unpaired electron in its valence shell.

The nitrogen atom has the following valence electronic configuration
2s2 2px1 2py1 2pz1. Since there is a double bond in nitrogen dioxide, we can expect it to have an unhybridised 2p orbital. So it undergoes sp2 hybridisation in which 2s2 2px1 2py1 are hybridised to form three sp2 hybrdised orbital, leaving unhybridised spz1 orbital. One of the three sp2 orbital would contain an electron pair while the remaining two are with an unpaired electron.

In nitrogen dioxide, the nitrogen atom forms a double covalent bond with a neutral oxygen atom and a dative covalent bond with another neutral oxygen atom.

The formation of double bond involves one of the sp2 hybridised orbital with single unpaired electron which head-on overlaps with 2px1 orbital of oxygen atom to form sigma bond. Also , it involves unhybridised 2pz1 orbital of which sideway overlaps with 2pz1 orbital of the oxygen atom to form pi bond.

The dative covalent bond is formed when the sp2 hybridsed orbital with paired electrons is donated to the empty 2pz orbital of oxygen atom.

The unpaired electron found in the nitrogen atom is in fact the sp2 hybridised orbital with single unpaired electron.

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

Nitrogen monoxide

Nitrogen monoxide is also a free radical. We can expect the nitrogen atom to have an unpaired electron in its valence shell.

The valence electronic configuration of nitrogen atom is the same, ie ,
2s2 2px1 spy1 2pz1. Since in nitrogen monoxide, there is a double covalent bond, we expect it to have an unhybridised 2p orbital. As a result, I believe that it undergoes sp2 hybridisation.
2s2 2px1 spy1 ---> three sp2 hybridised orbitals + one unhybridised 2pz1 orbital ... [ One of the three sp2 hybridised orbitals contain an electron pair while the other two contain only a single unpaired electron]

One of the sp2 hybridised orbital with unpaired electron is used to form sigma bond with the 2px1 orbital of oxygen atom while the unhybridised 2pz1 is used to form pi bond with the 2pz1 orbital of oxygen atom.

The remaining sp2 hybridised orbital with a paired electrons accounts as the lone pair found in NO while the other sp2 hybridised orbital with a single unpaired electron accounts for the free-radical properties.

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

Hopefully I didn't confuse any of you this time. I apologise for my ambiguous and fallacious replies earlier. There is no such thing as s hybridisation :)) (Brain malfunctioning)

And ChongKeat, I cannot find the intuition on how pumice works.

[Athersin]

for nitrogen monoxide, N=O, there is 2 lone pairs around the nitrogen atom rite?
So u mean that the 2 lone pairs around the N aton and the sigma bond between N and O atoms form the sp2 hybridised orbitals?

[markwongsk]

Athersin you happen to pick on one of the anomalies of nature =) There are several different school of thoughts on this molecule:

Lewis dot diagram:

In the lewis dot diagram for both atoms to fulfill the octet rule we would need a triple bond, between N and O but one of these bonds would be a coordinate bond (dative bond) where the electrons come from oxygen.

Some people argue that this is not the correct structure since for this particular structure Oxygen has a positive formal charge (3 from each bond pair and 2 from the lone pair) this is impossible since Oxygen is more electronegative than Nitrogen.

Then there are those who favor formal charge distribution: ie those who postulate that by nature molecules tend to use the structure which has the least magnitude of formal charge. In this scenario NO has a double bond, with N having a lone electron radical. Both atoms have 0 formal charge.

Then we have a combination of these two, which says that NO has a partial third bond. Because NO has a lone radical, electron from O tends to move to NO to complete the octet, yet Oxygen is more electronegative than N and so the electron is distributed.

A better interpretation would probably be using Molecular Orbital Theory. But that's not in Form 6 =)

I believe they won't ask you this on hybridization. because the shape is definitely linear.

Update: Okay my bad, didn't see the first page. Apparently it did came out. Well, I also don't know. But I'm making an educated guess that it's intermediate of sp and sp2... don't ask me what that is though, I've never even encountered it

[DonFrank]

for nitrogen monoxide, N=O, there is 2 lone pairs around the nitrogen atom rite?
So u mean that the 2 lone pairs around the N aton and the sigma bond between N and O atoms form the sp2 hybridised orbitals?

In nitrogen monoxide, there is only one lone pair in the valence shell of nitrogen atom. Because you see, the double bond consists of a sigma bond and a pi bond. The sigma bond takes 1 of nitroge's sp2 hybridised orbital and the pi bond takes the unhybridised 2pz1 orbital of nitrogen atom. So 2 out of 5 valence electrons of nitrogen is used up. leaving 3... 2 of it accounts for the lone pair (sp2 hybridised orbital with a paired electrons) and remaining one accounts for its free radical properties ( sp2 hybridised orbital with unpaired electron).

http://img40.imageshack.us/img40/9/picsv.png

*Correction; 2px1 orbitals of oxygen is in fact, sp2 hybridised orbital of oxygen.

Athersin you happen to pick on one of the anomalies of nature =) There are several different school of thoughts on this molecule:

Lewis dot diagram:

In the lewis dot diagram for both atoms to fulfill the octet rule we would need a triple bond, between N and O but one of these bonds would be a coordinate bond (dative bond) where the electrons come from oxygen.

Some people argue that this is not the correct structure since for this particular structure Oxygen has a positive formal charge (3 from each bond pair and 2 from the lone pair) this is impossible since Oxygen is more electronegative than Nitrogen.

Then there are those who favor formal charge distribution: ie those who postulate that by nature molecules tend to use the structure which has the least magnitude of formal charge. In this scenario NO has a double bond, with N having a lone electron radical. Both atoms have 0 formal charge.

Then we have a combination of these two, which says that NO has a partial third bond. Because NO has a lone radical, electron from O tends to move to NO to complete the octet, yet Oxygen is more electronegative than N and so the electron is distributed.

A better interpretation would probably be using Molecular Orbital Theory. But that's not in Form 6 =)

I believe they won't ask you this on hybridization. because the shape is definitely linear.

Update: Okay my bad, didn't see the first page. Apparently it did came out. Well, I also don't know. But I'm making an educated guess that it's intermediate of sp and sp2... don't ask me what that is though, I've never even encountered it

That partial third bond comes from the electron in oxygen? Is that the top gapped line?
http://img255.imageshack.us/img255/2669/2009117112315.png

I think lewis structure fails in depicting the bonding of structure involving atom with odd-number valence electrons

[Athersin]

In nitrogen monoxide, there is only one lone pair in the valence shell of nitrogen atom. Because you see, the double bond consists of a sigma bond and a pi bond. The sigma bond takes 1 of nitroge's sp2 hybridised orbital and the pi bond takes the unhybridised 2pz1 orbital of nitrogen atom. So 2 out of 5 valence electrons of nitrogen is used up. leaving 3... 2 of it accounts for the lone pair (sp2 hybridised orbital with a paired electrons) and remaining one accounts for its free radical properties ( sp2 hybridised orbital with unpaired electron).

that means the sp2 hybridisation account for the sigma bond formed by the N and O, the unpaired elctrons , as well as the lone pairs on N atom?

thanks mark for the clarity..

anomality always exist for odd number valance electrons, is it?

[DonFrank]

that means the sp2 hybridisation account for the sigma bond formed by the N and O, the unpaired elctrons , as well as the lone pairs on N atom?

thanks mark for the clarity..

anomality always exist for odd number valance electrons, is it?

Yes, exactly.

yup... so it is not a good idea to depict the bonding of molecules involving atom with odd number valence electrons using lewis structure. (due to incomplete octet)

[Athersin]

I thought only those orbitals that have been hybridised ( Mix with other orbitals) will be counted only. But then, N atom have one unpaired electron and one lone pairs, both of them do not hybridised, that is mixing with other orbitals to form bond , so why they are still counted?
then, should it mean that NH3 has sp3 hybridisation?
How about SO3 and SO4?

[DonFrank]

I thought only those orbitals that have been hybridised ( Mix with other orbitals) will be counted only. But then, N atom have one unpaired electron and one lone pairs, both of them do not hybridised, that is mixing with other orbitals to form bond , so why they are still counted?
then, should it mean that NH3 has sp3 hybridisation?
How about SO3 and SO4?

Those atoms that are not involved in bonding such as lone pair and unpaired electrons are in fact found in hybridised orbitals.

Yeah, N in NH3 uses sp3 hybridisation. The lone pair in the nitrogen atom is found within the sp3 hybridised orbitals.

S in SO3 is surrounded by 6 pair of electrons. S uses all its 6 valence electrons in the bonding, forming three double bonds. So we know that it must have three hybridised orbitals(three sigma bonds) and three unhybridised orbitals (three pi bonds). Therefore, it must have used sp2 hybridisation.

I assume that the SO4 you mention is with charges, SO4(2-).
S in SO4(2-) forms two double bond with 2 oxygen atoms respectively and two single bonds with another two uninegatively oxygen respectively.

For the two double bonds, S uses a total of two hybridised orbitals to form sigma bond and also two unhybridised orbitals to form the pi bonds. For the two single bonds, S uses two hybridised orbital to form two sigma bonds respectively in each single bonds mentioned.

S uses 4 hybridised orbitals and 2 unhybridised orbitals (6 valence electrons). Therefore, it must have undergone sp3 hybridisation.

[yeng]

I thought only those orbitals that have been hybridised ( Mix with other orbitals) will be counted only. But then, N atom have one unpaired electron and one lone pairs, both of them do not hybridised, that is mixing with other orbitals to form bond , so why they are still counted?
then, should it mean that NH3 has sp3 hybridisation?
How about SO3 and SO4?

The electron pair geometry - (type of hybridisation) determine the shape of the molecule. so there is sp3, sp2, sp hybridisation. this differs from the geometry of the molecule which is only determined by the visible molecules.


yes NH3 is sp3 hybridise - with 3 bonding pairs (bonded with H) and 1 lone pair (basically unpaired e- is called lone pair, therefore lonepair= unpaired e-. )

what do you mean counted? does it mean the shape of the molecule? the shape of the molecule is determined by the atoms not the hybridisation. lone pair is considered as invisible, therefore the shape of NH3 is pyramidal.

S have 6 valence electrons. SO3 is a trigonal planar molecule - sp2 hybridise and SO4 is a tetrahedral molecule. - sp3 hybridise.
do not forget S have empty d orbitals where the extra electrons can fill in the d orbital, thus it is possible for S to have more than 8 valence electron.

[DonFrank]

The electron pair geometry - (type of hybridisation) determine the shape of the molecule. so there is sp3, sp2, sp hybridisation. this differs from the geometry of the molecule which is only determined by the visible molecules.


yes NH3 is sp3 hybridise - with 3 bonding pairs (bonded with H) and 1 lone pair (basically unpaired e- is called lone pair, therefore lonepair= unpaired e-. )

what do you mean counted? does it mean the shape of the molecule? the shape of the molecule is determined by the atoms not the hybridisation. lone pair is considered as invisible, therefore the shape of NH3 is pyramidal.

S have 6 valence electrons. SO3 is a trigonal planar molecule - sp2 hybridise and SO4 is a tetrahedral molecule. - sp3 hybridise.
do not forget S have empty d orbitals where the extra electrons can fill in the d orbital, thus it is possible for S to have more than 8 valence electron.

isn't lone pair a valence electron pair which is not shared with other atoms?

[yeng]

yeah sorry is unpaired valance e-. u r right.

[Athersin]

sorry, i mistype it as s04, it is s02 that i am asking.

s02 is sp2 hybridised right?

O=S=O, i dont really know does there is any lone pairs or unpaired electrons around S atom, that is the main problem and how am i going to determine that?

S can have expanded structure , same with Phosphorus right?

PS: for the previous solved ques: international books do not lying or making mistakes..lesson learnt! haha

[yeng]

1. count the valence e- of each atom.
2. draw the lewis structure of the molecule by using octet rule.
3. u can see clearly what is the shape from the lewis structure u drawn.

[DonFrank]

sorry, i mistype it as s04, it is s02 that i am asking.

s02 is sp2 hybridised right?

O=S=O, i dont really know does there is any lone pairs or unpaired electrons around S atom, that is the main problem and how am i going to determine that?

S can have expanded structure , same with Phosphorus right?

PS: for the previous solved ques: international books do not lying or making mistakes..lesson learnt! haha

O=S=O
S forms 2 double bonds with 2 neutral oxygen atoms. We can expect that each oxygen atoms to share two of its electrons with S.

S itself has 6 valence electrons. Since it forms 2 double bonds, it must have got 2 unhybridised orbitals.

Look at its valence electronic configuration : 3s2 3px2 3py1 3pz1
To obtain 2 unhybridised orbitals with unpaired electrons (to form pi bonds), it must undergo sp hybridisation (hybridise 3s2 and 3px2) to form 2 sp hybridised orbitals. 1 of the sp hybridised orbitals will remain as the lone pair whle another (2 electrons with each e- used in each double bonds) will involve in head-on overlapping with the hybridised sp2 orbitals of oxygen.

*Oxygen undergoes sp2 hybridisation when it forms a double bond with other atom and sp3 hybridisation during formaton of single bond with other atom.

S has empty d-orbitals, so it can have expanded octet just like Phosphorus ( PCl5)...

The following is the way I usually use to predict molecular shape of the molecules.

Electrons that are shared
-------------------------
S - 6 electrons (for central atom, we list out its total valence e- of it)
2O - 4 electrons (for atoms bound to central atom, we list out total number of valence e- they (2 oxygen atoms) share with the central atom)
------------------------

To know how many bonding pairs are there,
we simply look at how many electrons from the peripheral atoms are shared, if there are 4 electrons being shared by 2 peripheral atoms, this means there will be 2 bonding pairs ( 4 / 2 = 2 ; in predicting molecular geometry, we consider single bond, double bond or triple bond as 1 bonding pair) ... [If there are 4 electrons being shared by 4 peripheral electrons, there will be 4 bonding pairs... eg, in CH4]

Total number of electrons pairs = (6+4) / 2 = 5
Number of lone pair = total no of electron pairs - electron pairs being shared
= 5 electrons pairs (total) - 4 electron pairs(shared electrons) = 1

So there are 2 bonding pairs and 1 lone pairs in SO2, its molecular shape is predicted to be bent or V shape.

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

[extell]

the best way to study org chem is to draw a mind map on paper and section the entire info in your brain. did that for alevels. worked.

[Sola]

Wonder why during esterification involving phenol and carboxylic acid have no reaction?why phenol only react with acyl chloride?i thought phenol is reactive enuf?

[zanyzephyr]

Wonder why during esterification involving phenol and carboxylic acid have no reaction?why phenol only react with acyl chloride?i thought phenol is reactive enuf?

correction:

Phenol is an OH-bonded to a benzene group. Reactivity of a phenol is low as the delocalized electrons in the benzene stabilizes the resonance ring structure, hence a normal reaction without suitable catalysts will not be reactive enough for the phenol to undergo esterification.

acyl chloride reacts with phenyl with a catalyst of FeCL3 or AlCL3, it wouldn't react with acyl chloride alone.

The chloride atom of acyl cl witll break up to form a temporary ion with Alcl3 to form alcl4-. as acl3 is a electron acceptor due to the partially filled orbital of alcl3.

hence, an electrophile will be formed, and electrophillic substitution will take place where the acyl group replaces the OH group.

Therefore, your statement was wrong. Phenol is not a reactive substance by itself. Please recheck your text book or reference book( i suggest Longman) again.

[dkpit]

Just want to ask if in stpm we are required to write [O] in equations involving oxidation or if we can leave it out?

[zanyzephyr]

Just want to ask if in stpm we are required to write [O] in equations involving oxidation or if we can leave it out?


no you can't leave the source of oxidation away, for example:


acidified hot KMnO4
C6H5CH3---------> C6H5COOH+ H2O


You must write the source of oxidation, and the presentation with [O] will be penalized.