Chemistry: Post your doubts here!

[Keshani Privinsha]

2.40
g of propan-2-ol were mixed with excess acidified potassium dichromate(
VI
). The reaction
mixture was then boiled under reflux for twenty minutes. The organic product was then collected
by distillation. The yield of product was 75.0%

 

[Djlucifer]

Um............I have my Chemistry exams this May/June 2017 session and have about 50 days in my hand so what do I exactly do to get the most of the marks as in an A and how do I plan my revision and stuff like that How do I do really good in Organic Chemistry in both Paper 1 and Paper 2 and some general tips and tricks for solving PApaer 1 which in my opinion is quite hard to some extent

 

[iSean97]

View attachment upload_2017-3-12_23-44-6.png

Hi there peeps, just wondering why is it not A. 3? I drew all the possible structure out.

 

[Error Syntax]

View attachment 61771

Hi there peeps, just wondering why is it not A. 3? I drew all the possible structure out.

Okay , yeah this one is tricky so basically what the examiner wants you to determine is the isomers but in benzene rings there is a hidden isomer which is considered to be where the chlorine atoms are attached to double carbon bonds, which in your diagram is the 4th one and if you ask how it differs from your first well its the difference in single bonds and double bonds which in benzene is something unique.

 

[techgeek]

Somehow I can't get any of the answers provided in question 5.

 

[Error Syntax]

here you go

 

[Amir Anwar]

what will be the product of HOCH2CH(OH)CO2H+CH3CO2H
m/j/09/21/q4 b2

 

[darks]

what will be the product of HOCH2CH(OH)CO2H+CH3CO2H
m/j/09/21/q4 b2

CH3COOCH2CH(OCOCH3)COOH ????

 

[Thought blocker]

CH3COOCH2CH(OCOCH3)COOH ????

Yes.

 

[student2]

I assume this is a simple concept. To know whether an atom donates a pair of electrons or shares them to expand it's octet, you have to think about each atom and how many electrons it requires to complete it's octet.
the atom that donates it's pair of electrons has to have at least one pair of lone pair of electrons after it has formed the bond.
also you have to know, that in a dative bond, the electron donor atom donates electrons in pairs. (in the case you mentioned, 3 electrons donated are not in pairs. hence in this case chlorine has to share those 3 electrons with 3 fluorine atoms to expand it's octet.) if an atom has those qualities metioned then in this case it will donate a lone pair instead of expanding it's octet.

you can look at other similar example to help you,

• like Al2Cl6, ( in this case there is no octet expansion taking place)

here each aluminium atom is covalently bonded with 3 chlorine atoms , while at the same time it's accepting a pair of electrons from a forth chlorine atom,,, take a look at the fourth chlorine atom, (as mentioned before the atom donating the pair of electrons has to have at least one pair of electrons after forming the bond, at the same time it donates the electrons in pairs.)
I hope this answers your question

z ths still dere in 2016 syllabus? i cnt fynd it in 2015 syllabus either...

 

[Holmes]

9701/12/O/N/10
October 2010 paper 1 : Q8 ,Q11 , Q28, Q34, Q39.
Help me please!!
Thanks in advance

 

[techgeek]

9701/12/O/N/10
October 2010 paper 1 : Q8 ,Q11 , Q28, Q34, Q39.
Help me please!!
Thanks in advance

Q8:
It's D. First IE of calcium = 590
Second IE of calcium = 1150

To convert solid Ca into gas you need 177 kJmol-1 energy, and then to convert it to 2+ ions you need 590+1150 energy. Then for hydrating the ions the energy change is 1565kjmol-1. So total energy change = 177+590+1150-1565 = +352 kJmol-1

Q11: Initial concentration of X: 0.50
concentration of X during equilibrium: 0.25
Initial concentration of Y : 0
concentration of Y during equilibrium: since 1 mol of X produces 2 mol so conc of Y= 0.25x2 = 0.50
Kc= [Y}^2/[X] =0.50^2/0.25 =1
So it's C


Q28: In skeletal formula, we don't display Hydrogen atoms, so A is incorrect.
C shows 5 carbon atoms while D shows 3, but the repeat unit shows there should be only 2 carbons with one florine which is in B only.

Q34: The reactions will be
Mg + 2 CH3COOH <=> Mg(CH3COO)2+ H2(g) (since ethanoid acid is a weak acid, it will be in equilibrium)
Mg + H2SO4 >> MgSO4 + H2
1 After 2 minutes, the sulfuric acid is at a higher temperature than the ethanoic acid.
2 After 2 minutes, the sulfuric acid has produced more gas than the ethanoic acid
Sulfuric acid is a stronger oxidizing agent and a diprotic acid while ethaonic acid is monoprotic, that means it will give off more hydrogen ions per unit time than ethanoic acid so more hydrogen gas will be produced. Since bond formation is exothermic, sulfuric acid will also be at a higher temperature
3.After 20 minutes, the sulfuric acid has produced more gas than the ethanoic acid.
The reaction between ethanoic acid and magnesium is in equilibrium, so it never reaches completion.
So the answer is A.

Q39: It's just calculation, find Mr for each of the reactants and products
no. of moles of reactants/no. of moles of products x 100%
you'll find they all give 62% yield so the answer is A.

that's it

 

[amina1300]

A space shuttle’s upward thrust came from the following reaction between aluminium and ammonium perchlorate.
10Al + 6NH4ClO4 → 4Al 2O3 + 2AlCl 3 + 12H2O + 3N2
Which statements about this reaction are correct?
1 Aluminium is oxidised. 2 Chlorine is reduced. 3 Nitrogen is oxidised.

 

[Thought blocker]

A space shuttle’s upward thrust came from the following reaction between aluminium and ammonium perchlorate.
10Al + 6NH4ClO4 → 4Al 2O3 + 2AlCl 3 + 12H2O + 3N2
Which statements about this reaction are correct?
1 Aluminium is oxidised. 2 Chlorine is reduced. 3 Nitrogen is oxidised.

 

[Holmes]

A space shuttle’s upward thrust came from the following reaction between aluminium and ammonium perchlorate.
10Al + 6NH4ClO4 → 4Al 2O3 + 2AlCl 3 + 12H2O + 3N2
Which statements about this reaction are correct?
1 Aluminium is oxidised. 2 Chlorine is reduced. 3 Nitrogen is oxidised.

Option D is only correct. i.e.
1 Aluminium is oxidised. (only)
why
If we look at the oxidation state of Al than we can clearly see that Al is oxidised from 0 to +3
while chlorine's oxidation state remains the same.i.e. -1
And Nitrogen is reduced from +5 to 0 while statement 3 says the opposite.
so answer is D

Hope it helped.

 

[Lee Qian Yi]

can explain why answer is C?

 

[Holmes]

can explain why answer is C?View attachment 61871

KMnO4 is a famous oxidising agent. Using COLD KMnO4 will break "C=C" and add an -OH group on each carbon atom. Now count the total number of hydroxy groups present. Two are attached with each Carbon and one already attached to your absolute LEFT hand side (in the diagram ).
So this narrows your choice and you are restricted to "C" and "D" only.
When hot ,concentrated acidified KMnO4 is used then this reagent will cause the "C=C" to break, thus making ketone. In this way ONE of the 6 - membered is broken. Now count the remaining 6-membered rings, and you will find that there are ONLY TWO (2) remaining 6- membered rings.
So your answer is "C".
I hope I helped you.

 

[Lee Qian Yi]

ok thx , i know already.

 

[Amir Anwar]

Yes.

how can u solve it

 

[Thought blocker]

how can u solve it

solve what?