Edexcel - Chemistry Unit 2 - 4 June 2013

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I got the answer for this (A), but the question overall confused me. [from June 2011]
Shouldn't a secondary alcohol be heated under reflux to give a ketone? Or is this just a case of choosing the "more correct answer"? :tongue:

Reply 301

username1097440

Original post by SKK94

what is the question about :confused:

Reply 302

posthumus

Original post by SKK94

Nope you can use both... it's just that heat under reflux is typically used. The distillation part is not needed as it's not further oxidized but you can still use the apparatus :smile:

Original post by JRN-95

what is the question about :confused:

Oxidation of Primary & Secondary alcohols... if you have not come across that yet, then I suggest you get reading my friend !

Reply 303

posthumus

Original post by Mollymod

Oooh I have high hopes now :biggrin:

I wrote "To remove any acid from the food" then while the teachers were collecting the papers... I decided to open my paper up & just write "salt" as well for that question :tongue:

So hopefully I get 1 mark at least :frown:

& I've firmed Computer Science at King's College London :smile:

& yourself? Pharmacy ? ... something chemistry related definitely, judging from your TSR name :biggrin:

Reply 304

Jayqwe

When answering questions about the halgenoalkane reactions such as the substitution and elimination ones, and the question says something like "How do you turn a halgenoalkane into a alcohol?" Would you answer by saying reflux with an aqueous alkali hydroxide or just alkali?

Basically do you write alkali hydroxide or alkali?

Reply 305

Mollymod

Original post by Jayqwe

Well I'd be more specific and say KOH or NaOH just to be safe, so in respect that's an alkaline hydroxide but they often want specifics.

Original post by posthumus

Oooh I have high hopes now :biggrin:

I wrote "To remove any acid from the food" then while the teachers were collecting the papers... I decided to open my paper up & just write "salt" as well for that question :tongue:

So hopefully I get 1 mark at least :frown:

& I've firmed Computer Science at King's College London :smile:

& yourself? Pharmacy ? ... something chemistry related definitely, judging from your TSR name :biggrin:

Yeah I'm pretty sure that as long as you have a plausible reason and back it up properly using Chemistry you should be fine.
Biochemistry with a year abroad at Essex :smile:

(edited 10 years ago)

Reply 306

Jayqwe

Original post by Mollymod

Well I'd be more specific and say KOH or NaOH just to be safe, so in respect that's an alkaline hydroxide but they often want specifics.

Thank you

Reply 307

pineapple78

10. 20 cm3 of sulfuric acid, concentration 0.25 mol dm–3, was neutralized in a titration with barium hydroxide, concentration 0.50 mol dm–3. The equation for the reaction is
Ba(OH)2(aq) + H2SO4(aq) → BaSO4(s) + 2H2O(l)

(a) The volume of barium hydroxide required was (1)
A 10 cm3
B 20 cm3
C 25 cm3
D 40 cm3

help anyone?

Reply 308

posthumus

Original post by pineapple78

Is the answer A ? If so...

The ratio between reaction of sulfuric acid & barium hydroxide is 1:1 ... therefore they must both equal the same amount of moles.

moles = volume x concentration

therefore... mole of sulfuric acid = 0.02 x 0.25 = 0.005

there must also be 0.005 moles of barium hydroxide, therefore:

volume= moles/ concentration = 0.005/ 0.5 = 0.01

Woops my bad :tongue:

they want it in cm^3... here we have our answer in dm^3

1dm^3 = 1000 cm^3
so...

0.01 x 1000 = 10 cm^3

Reply 309

saifulahmed

January 2012, question 20(a)(iv).How do we find the concentration?The question is really big, so i'm not posting it!

Reply 310

Gnome :)

Original post by saifulahmed

January 2012, question 20(a)(iv).How do we find the concentration?The question is really big, so i'm not posting it!

This is what I've come up with, let me know if it makes sense or not :smile:

SCAN0023.JPG294.6KB

Reply 311

SKK94

Original post by posthumus

Nope you can use both... it's just that heat under reflux is typically used. The distillation part is not needed as it's not further oxidized but you can still use the apparatus :smile:

Oh! Thanks :smile:

Reply 312

zyxwvuts

I am terrified about this exam. I'm so scared it's going to be weird and difficult, not forgetting that the grade boundaries seem to be slowly climbing up :frown:

Reply 313

Miri321

Hey, does anyone have any advice as to how to answer a section C question successfully - there's always that 6 mark question, and I always get it wrong.

Reply 314

Jayqwe

Section c always has a random topic each time so that is hard say, also do we need to know how to turn an amine, alcohol and an alkene back to a halgenoalkane, or do we just need to know the forward reactions?

Posted from TSR Mobile

Reply 315

username1097440

Original post by posthumus

Nope you can use both... it's just that heat under reflux is typically used. The distillation part is not needed as it's not further oxidized but you can still use the apparatus :smile:

Oxidation of Primary & Secondary alcohols... if you have not come across that yet, then I suggest you get reading my friend !

i understand primary, secondary and tertiary alcohols and their structure what i dont understand is when they say oxidised how can you work out what product is formed without doing the oxidation numbers?

Reply 316

Priya08

Hii
does anyone know apart from reflux and distilation, what other diagrams do we need to know?

and does anyone know what diagram to draw for this question:

Jan 2011 - Q13

Draw a diagram of a seperating funnel containing the seperated layers! label the hydrocarbon and state its colour

Reply 317

scientific222

Original post by JRN-95

You don't do any oxidation numbers. When a primary alcohol is oxidised it always forms an aldehyde/carboxylic acid, when a secondary alcohol is oxidised it always forms a ketone and tertiary alcohols cannot be oxidised. For the product formed, draw out the structure of the alcohol and if it is a primary alcohol being oxidised into an aldehyde, you remove the -OH part and add a double bond =O and a -H to the carbon that was carrying the -OH. Similarly for secondary alcohols ketones, you remove the -OH and add a double bond =O to the carbon carrying the -OH.
Read this: http://www.chemguide.co.uk/organicprops/alcohols/oxidation.html

Reply 318

scientific222