Advanced Higher Chemistry 2012-2013

Reply 980

TheFOMaster

13

Original post by Zahra Tasleen

Don't understand sorry

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He means to find the major product you have to apply markovnikovs rule :smile:

(I think)

Reply 981

SanahAhmed

Original post by Zahra Tasleen

Don't understand sorry

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Well the answer is D. It cannot be C because but-2-ene, when Cl or I add on would only add onto Carbon 2 and 3 as this is where the double bond breaks, so C cannot be an answer. Also option A and B imply that only Cl or only I can add however they would compete as depicted in option D.

Reply 982

I am Ace

4

Original post by TheFOMaster

He means to find the major product you have to apply markovnikovs rule :smile:

(I think)

Original post by Zahra Tasleen

Don't understand sorry

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I do!

But I didn't realise it was asking for both the products.

OK, so Alkene Halogenation forms 2 products each with an Alkene attached to 2 Halogens

So where will these Halogens likely go?

Hlogens are Electrophiles remember and the double bond is electron rich

Reply 983

Lala:

Original post by Zahra Tasleen

I'm also stuck on these 2 thanks in advance ImageUploadedByStudent Room1369754432.136950.jpg

32 and 33

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In 33 -You need to know that benzene is planar The answer can't be A or B. The benzene in D is substituted with an alkyl group so it isn't D either. C is a benzene ring substituted with a chlorine atom-so the answer is C

Reply 984

SanahAhmed

Original post by Zahra Tasleen

I'm also stuck on these 2 thanks in advance ImageUploadedByStudent Room1369754432.136950.jpg

32 and 33

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Question 32: Sodium OH would react with C as it would provide an NaOH to break the ester linkage open and form sodium ethanoate and ethanol. Option D is an ether and you know they don't react. Option B is an alkene however it would result in an addition not a complete reaction, also option A cannot react as it already contains a OH group.

Question 33: We know that benzene is planar, so option C is correct as a H is only replaced by a Cl. However (just in case you ask) option D would be incorrect as the CH₃ would mean the molecule is no longer planar.

Does that help, or shall I come round to your house :tongue:

?????

(edited 10 years ago)

Reply 985

Zahra Tasleen

Original post by SanahAhmed

Question 32: Sodium OH would react with C as it would provide an NaOH to break the ester linkage open and form sodium ethanoate and ethanol. Option D is an ether and you know they don't react. Option B is an alkene however it would result in an addition not a complete reaction, also option A cannot react as it already contains a OH group.

Question 33: We know that benzene is planar, so option C is correct as a H is only replaced by a Cl. However (just in case you ask) option D would be incorrect as the CH₃ would mean the molecule is no longer planar.

Does that help, or shall I come round to your house :tongue:

?????

No that's fine

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Reply 986

stephanieahahah

see how for question 2(a) and (b) i'm confused at what values/equation to use to work out the answer?
2a is -257.5kJ mol-1 and b is 199 kJ mol-1

Reply 987

FeelingofSuccess

Original post by laurenmackessack

Your lucky i have advanced higher geography the day before chemistry! haha good luck for french

Original post by emma_1995

YES. I've got French tomorrow as well (higher) and have been basically neglecting it since chemistry is the AH.

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Oh, the day before is brutal! Good luck!

Yup, I have Higher french too and have been focussing on that for a while.. Haha! :tongue:

Let's hope we'll all be okay! Good luck to y'both :smile:

Reply 988

deedee123

16

Original post by FeelingofSuccess

Anyone else finding it annoyingly difficult to revise for 2 exams when they are so close together? I have French tomorrow as well as this Chemistry exam!

im the exact same, i have geography on thursday and have done no studying at all for it :tongue:

Reply 989

Chemistry Help

Can someone please help me in these questions?
Answers are in bold

Which of the following alcohols would have
the greatest entropy at 90 ºC?
A Propan-1-ol
B Propan-2-ol
C Butan-1-ol
D Butan-2-ol

The activation energies for the reactions
(1) H2 (g) + I2 (g) → 2HI(g)
(2) 2HI(g) → H2 (g) + I2 (g)

are 165 kJ and 179 kJ respectively.
The enthalpy change for reaction (2) is

A –14 kJ
B +14 kJ
C –344 kJ
D +344 kJ.

Thanks in advance :smile:

Reply 990

I am Ace

4

Why is the carbocation that is the most stable, the one that produces the major product?

Shouldn't the least stable, be the one that needs a nucleophile the most?

Reply 991

emma_1995

Original post by FeelingofSuccess

Oh, the day before is brutal! Good luck!

Yup, I have Higher french too and have been focussing on that for a while.. Haha! :tongue:

Let's hope we'll all be okay! Good luck to y'both :smile:

Thanks

good luck to you too :smile:

can't wait till Friday afternoon when I will be free :smile:

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Reply 992

BaconMan

Original post by stephanieahahah

see how for question 2(a) and (b) i'm confused at what values/equation to use to work out the answer?
2a is -257.5kJ mol-1 and b is 199 kJ mol-1

Draw out the structure of ethanol and count how many C-H bonds there are plus the O-H bond. Look up the mean bond enthalpies in the data booklet for these bonds. Times the value for C-H bond with how many there are and add the value for 0-H. This should give you the enthalpy of formation of this compound as it is the sum of all its component bonds :smile:

Reply 993

I am Ace

4

Original post by MOHSINM31

Can someone help me with section A Q10 2012? I know how to use the equilibrium formula but I'm struggling to figure out the mole/concentration to put in the formula... Please help me!

Posting a link to the question increases your chances of being answered

Reply 994

MOHSINM31

I deleted my last post(Q10-2012) as I managed to figured it out.

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Reply 995

MOHSINM31

Original post by I am Ace

Posting a link to the question increases your chances of being answered

I'll keep that in mind for next time, I know how to do it now.

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Reply 996

I am Ace

4

Original post by MOHSINM31

I deleted my last post(Q10-2012) as I managed to figured it out.

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You don't have to delete it, buddy, you can simply edit it saying

EDIT: got it now

Reply 997

stephanieahahah

Original post by BaconMan

Draw out the structure of ethanol and count how many C-H bonds there are plus the O-H bond. Look up the mean bond enthalpies in the data booklet for these bonds. Times the value for C-H bond with how many there are and add the value for 0-H. This should give you the enthalpy of formation of this compound as it is the sum of all its component bonds :smile:

thank you

i've got it now, i didn't realise you have to take the sublimation value of carbon