Chemistry: Post your doubts here!

[Browny]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w04_qp_2.pdf

Can anyone explain why in Q3)b)ii) why they are saying in the marking scheme that bonding changes from covalent to ionic?
Because my opinion is that HCl is ionic.

 

[BeBeskii108]

I've done more than 10 Paper4's and I still get around 50-60. Any suggestions ? Those papers seem to vary from each other a lot. Very few straightforward answers. Can somebody please give a link of organic chemistry reactions required for A2 ?

 

[AbbbbY]

While we draw chiral compounds is there any specific rule as to which (CH3/OH/COOH etc) will be coming out of page and to which will be going into page?

It's a 3D structure so really, it makes no difference.

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w04_qp_2.pdf

Can anyone explain why in Q3)b)ii) why they are saying in the marking scheme that bonding changes from covalent to ionic?
Because my opinion is that HCl is ionic.

HCl IONIC!???! Good Lord!

2 days to p2 mate! HCl isn't ionic, it's covalent. Cl has 7 outer electrons, H has a single electron. How can they ever be ionic.

It shows ionic character in water because it's polar-covalent. HCl + H2O -> H3O+ + Cl- [thus an acid.]

 

[robinhoodmustafa]

HCl IONIC!???! Good Lord!

2 days to p2 mate! HCl isn't ionic, it's covalent. Cl has 7 outer electrons, H has a single electron. How can they ever be ionic.

It shows ionic character in water because it's polar-covalent. HCl + H2O -> H3O+ + Cl- [thus an acid.]

If its ionic in water than it must show a base nature so why have to called the product an acid?

 

[AbbbbY]

If its ionic in water than it must show a base nature so why have to called the product an acid?

What?

 

[robinhoodmustafa]

What?

Ionic compound are basic and Covalent ones are acidic. so how come to wrote 'thus an acid'

 

[AbbbbY]

Ionic compound are basic and Covalent ones are acidic. so how come to wrote 'thus an acid'

Firstly that is a very, very crude and rudimentary method to define acids and bases.

Secondly, ionic BEHAVIOR isn't the same as ionic-compound. Lets say someone is a man and behaves like a woman. He wont be a woman. Just a man who behaves like one. Get it?

Likewise, due to the H-Cl polarity, the bond with H2O present can from hydronium ion and chloride ion. (H3O+ and Cl-) (can also write it as H+ and Cl- but that wouldn't make sense further on). H3O+ is a proton donor, and thus classified as an acid.

 

[AbbbbY]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
can someone PLEASEEEE explain question 5b(iv)?!

Oh this question has given me a lot of trouble lately.

I went with (21/1)*(1/9) = 21/9 = 7/3


My explanation was something like 21/1 = ratio of rate of tertiary to primary H. 1/9 = ratio of tertiary to primary H present. Ratio of J/K = 21/1 * 1/9 = 21/9 = 7/3

J= Tertiary Hydrogen replaced.
K= Primary Hydrogen replaced.

 

[Logitech]

Guys I have a doubt. Now if we have an alkene, for example but-2-ene, when we react it with K2Cr2O7 it'll form CH3CH(OH)CH(OH)CH3 right? But what if we have an alkene with double bond at 2 different places like hex-2,5-ene, if we react this with K2Cr2O7 will a diol occur in both places?
Another question is when they ask for a reagent and the answer is potassium dichromate, should I write K2Cr2O7 or Cr2O7 2- or Cr2O7/H+ ?

 

[robinhoodmustafa]

Guys I have a doubt. Now if we have an alkene, for example but-2-ene, when we react it with K2Cr2O7 it'll form CH3CH(OH)CH(OH)CH3 right? But what if we have an alkene with double bond at 2 different places like hex-2,5-ene, if we react this with K2Cr2O7 will a diol occur in both places?
Another question is when they ask for a reagent and the answer is potassium dichromate, should I write K2Cr2O7 or Cr2O7 2- or Cr2O7/H+ ?

Yes. Both double bond with be broken and replace by OH on 5th and 2nd carbon bond. try drawing displayed formula.it really helps
and Cr2O7/H+ is liked by examiner. Mostly this is written in MS but K2Cr2O7 wont loose u marks.

 

[itallion stallion]

Can someone plz help me with this
http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
Q5biv
Thanks!

 

[AbbbbY]

Yes. Both double bond with be broken and replace by OH on 5th and 2nd carbon bond. try drawing displayed formula.it really helps
and Cr2O7/H+ is liked by examiner. Mostly this is written in MS but K2Cr2O7 wont loose u marks.

It will if you don't write ACIDIFIED.

Acidified K2Cr2O7 or (Cr2O7)2-/H+

 

[AbbbbY]

Can someone plz help me with this
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_43.pdf
Q5biv
Thanks!

Oh this question has given me a lot of trouble lately.

I went with (21/1)*(1/9) = 21/9 = 7/3


My explanation was something like 21/1 = ratio of rate of tertiary to primary H. 1/9 = ratio of tertiary to primary H present. Ratio of J/K = 21/1 * 1/9 = 21/9 = 7/3

J= Tertiary Hydrogen replaced.
K= Primary Hydrogen replaced.

 

[Suchal Riaz]

It will if you don't write ACIDIFIED.

Acidified K2Cr2O7 or (Cr2O7)2-/H+

if u don't write acidified or H+ or H2SO4 then u will not get marks.

 

[AbbbbY]

if u don't write acidified or H+ or H2SO4 then u will not get marks.

....which is exactly what I stated up there.

 

[Namehere]

Guys I have a doubt. Now if we have an alkene, for example but-2-ene, when we react it with K2Cr2O7 it'll form CH3CH(OH)CH(OH)CH3 right? But what if we have an alkene with double bond at 2 different places like hex-2,5-ene, if we react this with K2Cr2O7 will a diol occur in both places?
Another question is when they ask for a reagent and the answer is potassium dichromate, should I write K2Cr2O7 or Cr2O7 2- or Cr2O7/H+ ?

Isn´t this supposed to be with cold dilute acidifed KMnO4?

 

[itallion stallion]

From where have u brought this 9?

 

[AbbbbY]

Isn´t this supposed to be with cold dilute acidifed KMnO4?

Exactly what I was thinking. Was trying to jog my memory and see where in alkene reactions K2Cr2O7 was used. All I can remember is Cold dil KMnO4 -> Di-ol, hot conc -> cleavage.

 

[robinhoodmustafa]

Isn´t this supposed to be with cold dilute acidifed KMnO4?

Dilute Cold* KMNO4