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Chemistry Unit 2 Edexcel, Exam- 23rd May 2012

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Just a rubbish exam to be honest, about 1/4 of the syllabus was on the test and well it wasn't well worded either.

For anyone having problems with the concepts of Redox reactions take a look at: http://www.khanacademy.org/science/chemistry/v/redox-reactions :smile:

i actually thought it was ok.. i hope so

Reply 121

phaseshift

Original post by avipsita

i actually thought it was ok.. i hope so

I'm sure you did fine. :smile:

Right.. 2012 question paper or an older one?

Reply 122

AAFK

HHHHHHEY, can anyone please teme what's MORE likely to come? :s-smilie:

Reply 123

Arusa01

Original post by AAFK

HHHHHHEY, can anyone please teme what's MORE likely to come? :s-smilie:

Titrations they come up all the time - make sure you know how to do them. :smile:

Reply 124

Richee

Original post by The_New_Guy

Guys,
Do we have to know about the mechanisms of nucleophilic substitution of halogenoalkane as SN1 and SN2?

It is better if you learn it. Our teacher also taught us about it

Reply 125

jadez_94

Original post by LeaX

I've been looking through this thread and saw people mention Sn1 and Sn2 reactions? I've never even heard of these, could someone possibly explain them to me please?

SN stands for substitution nucleophilic. the number stands for the numbr of reactants involved at the start.

halogenoalkanes with OH- ions (from an alkali for example) to form an alcohol but the way in which the reaction occurs is different for primary and tertiary halogenoalkanes.

for primary halogenoalkanes they go through SN2 reaction. 2 because both 2 reactants are involved in the first step of the reaction (yno the arrow from the OH- to the C(delta +ve), and then an arrow from the bond between the C(delta +ve) to the Cl). This reaction is fast and happens in one continuous step. (there is a transition state where the C is attached to both OH and Cl and has a -ve charge, the OH comes in and Cl leaves in one motion) and you get your alcohol an a halide.

Then you have tertiary alcohols...these go through SN1 reactions (1 because 1 species/molecule is involved in the first step). Basically the OH- ion cannot initally attack the tertiary halogenoalkane because of sterical hinderance (the methyl groups are very bulky and there is no room for attack. so the chlorine has to leave first) the first step (the chlorine leaving) is very slow. the intermediate is a carbocation with the 3 methyl groups attached to a C and a positive on the Carbon. the next step is fast as the OH- comes in so create you tertiary alcohol.

and that's basically it, i got an exam question asking for the intermediate/transition stages in Jan 2012 Q21 aii) they want you draw it..simple enough but explain to for 4 marks.
2 marks for the drawing...easy marks but for the primary remember to put the -ve on the transition state because carbon doesnt really have 5 bonds so you do dotted lines. then the other 2 marks were for saying:
-that the tertiary carbocation more stable than the primary carbocation.
-the methyl groups stabalise the charge of the carbocation (through positive inductive effect)<dont worry about that i didnt know that "inductive effect"
-steric hinderance (by methyl groups) inhibits the formation of (trigonal bipyramidal) transition state/attack by nucleophile with tertiary compound
-steric hinderance is less with the primary halogenoalkane/more with the tertiary halogenoalkane

hope this helps x

Reply 126

verdikt

Can anyone tell me why ethoxyethane is insoluble in water? Yes, I know it can't form hydrogen bonds with water, but look at the molecule. It is C2H5-O-C2H5. Can't that O form hydrogen bonds with H2O?

Reply 127

Bord3r

Original post by verdikt

Can anyone tell me why ethoxyethane is insoluble in water? Yes, I know it can't form hydrogen bonds with water, but look at the molecule. It is C2H5-O-C2H5. Can't that O form hydrogen bonds with H2O?

I think it can't form hydrogen bonds because the Hydrogen is bonded to Carbon atoms only so the bonds aren't polar enough for H-Bonds to form unlike H20 where the H-O bond is much more polar.

Reply 128

oli_G

Original post by Blob2491

What are you guys getting in past papers?!

85% - 90% raw generally. I tend to get close to full marks on section A and B but then lose marks on section C :angry:

Reply 129

Bord3r

Original post by oli_G

85% - 90% raw generally. I tend to get close to full marks on section A and B but then lose marks on section C :angry:

Section C really frustrates me, especially the "suggest" questions.

Reply 130

salvequamtu

I don't get what the difference is between elimination and substitution reactions?? :frown:

Of Halogenoalkanes....
I mean they both require neuclophile OH from KOH
But for elimination is concentrated KOH used??? :s
& For substitution reactions, a dilute KOH solution is used ...?

Reply 131

im so fresh

need 100/120 in this module -.-

Reply 132

verdikt

Original post by Bord3r

I think it can't form hydrogen bonds because the Hydrogen is bonded to Carbon atoms only so the bonds aren't polar enough for H-Bonds to form unlike H20 where the H-O bond is much more polar.

Hmm, that's what I thought. I just realised how O may not be so electronegative, even though this is highly unlikely to come.

Reply 133

oli_G

Original post by AmrinderRai

I don't get what the difference is between elimination and substitution reactions?? :frown:

Of Halogenoalkanes....
I mean they both require neuclophile OH from KOH
But for elimination is concentrated KOH used??? :s
& For substitution reactions, a dilute KOH solution is used ...?

Elimination and substitution are two very different types of reactions and both of these occur occur when a halogenoalkane is mixed with KOH - however the extent to which elimination occurs or vice versa is determined by the reaction conditions.

If the halogenoalkane is mixed with aqueous KOH and heated under reflux substitution dominates - where the halogen in the haloalkane is replaced by the hydroxyl group of KOH as such:

CH_3CH_2X + KOH --> CH_3CH_2OH + KX

If the halogenoalkane is mixed with alcoholic KOH and heated under reflux elimination dominates - where the hydroxyl group acts as a nucleophile, taking a hydrogen from a carbon atom adjacent to the one in the C-X bond. This causes a C=C bond to form, and the halogen group leaves as it is the strongest leaving group. The reaction is as such:

CH_3CH_2X + KOH ---> CH_2CH_2 + H_2O + KX

Note that elimination and substitution occurs in both of these reactions.

Read through this page if you're still confused/ want more details: http://www.chemguide.co.uk/organicprops/haloalkanes/hydroxide.html

(edited 11 years ago)

Reply 134

EffKayy

http://www.scriblink.com/index.jsp?act=phome&roomid=330&KEY=26B042D56446C95EC2B78CD5E3471C56

Come revise with us

Reply 135

oli_G

Original post by EffKayy

http://www.scriblink.com/index.jsp?act=phome&roomid=330&KEY=26B042D56446C95EC2B78CD5E3471C56

Come revise with us

Deja vu

Reply 136

cisne

anybody have marking schemes for unit 2 old syllabus?

Reply 137

Blob2491

Original post by oli_G

85% - 90% raw generally. I tend to get close to full marks on section A and B but then lose marks on section C :angry:

Wow...tips on getting an A?

Reply 138

avipsita

Original post by im so fresh

need 100/120 in this module -.-

what did u get in the last module??

Reply 139

delphinas

Just curious, does anyone have all in one notes? Like a summary of everything ? Or maybe according to specification? Would be nice if someone would share it :} Coz most of the notes are incomplete :frown:

So scared of this exam, my third time lol!