Chemistry: Post your doubts here!

[AbbbbY]

MCQ QUESTIONS PLEASE HELP

Q6,17,20,22,24,25,26,27,28,29,38
With explanations, if you can help me with even one question, ty

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf

6:
I'd have picked D.

It's between A and D for me. I'll take D because you're getting rid of the double bond so 120 -> significantly lesser.

In A, and Alcohol will be formed but I'm not sure what the bond angles are. Should be lesser on one of the bonds due to the O lone pairs. Still I'd go for D because of the double bond -> single bond.
Edit: I just figured, it's the central carbon atom bond that we need to see not the O-H bond so A is out. A angles are all 109.5, the ones that matter.
17:
D

A- Happening.
B- NH3 -> NH4
C- (NH4)+ and (SO4)2-
D- You're just left with D. If you're unsure and want to waste more time, SO3 s = +6 final S = (-2+S-8)=0 so S= +6 so no redox.

22-
Free Radical Substitution
D

You're subtituting the H with Cl. Remember the very first reaction? FRSR of Alkane? That's what is happening. Don't let the acid part fool you it's the alkane part that's reacting here.

24-
D
Nucleophilic Substitution so both are nucleophiles

25-
B
A gives you propene
B gives you 2 bromo propane
(Remember, Halogenoalkanes, if theres a reaction that's not under ethanolic condition, it'll form an alcohol)
C gives you propene
D Again gives you propene (dehydration of the alcohol)


26:
Oxidation of a primary alcohol can give you Aldehydes and Carboxylic Acids.

27:
B
Ethanol reacting with Butanoic Acid so Ethyl Butanoate

28:
Will get back to you tomorrow.

29:
D

I can't explain this in words. Just draw it out and see where the double bonds would go. Adjacent Carbons in this case. let me know if you dont get it and I'll draw them.

 

[Namehere]

MCQ QUESTIONS PLEASE HELP

Q6,17,20,22,24,25,26,27,28,29,38
With explanations, if you can help me with even one question, ty

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s09_qp_1.pdf

Q20:

B- 2^3 = 6. If you know about statistics this is a bit easier. There are 2 possibilites around each double bond, either cis or trans. You have 3 double bonds in which this cis-trans combination can occur, hence it is 2 x 2 x 2 aka 2^3 = 6 (2 combinations for first double bond, 2 for second, 2 for third.)

Q38: B - 1 & 2 are correct. If you look at the beginning of the molecule from the left you can see that KMnO4 will break the double bond, and since the C attached to the double bond is a tertiary carbon atom (since its attached to 3 carbon atoms), it will yield, amongst other products, a ketone - in this case propanone. 2 is also correct because halogens can add across double bonds via the electrophilic addition mechanism.

Hope it helps.

 

[Namehere]

Abbbbby... I have researched a bit for Q28, and this is what I found: http://en.wikipedia.org/wiki/File:Carvone_oxidation.png

Which I don´t get, plus, that product has 8 carbons and the question said X has 9 carbons. If you do understand please explain to me.

With regards to your answer, I dont get the cleavage you did to the double bond in the cyclic compound, the part where you opened it - I get the aldehyde part of where you opened it, but what about the other carbon atom in the double bond (the one next to the CH3 branch), its a tertiary carbon atom, shouldn´t you get a ketone there?

 

[AbbbbY]

Abbbbby... I have researched a bit for Q28, and this is what I found: http://en.wikipedia.org/wiki/File:Carvone_oxidation.png

Which I don´t get, plus, that product has 8 carbons and the question said X has 9 carbons. If you do understand please explain to me.

With regards to your answer, I dont get the cleavage you did to the double bond in the cyclic compound, the part where you opened it - I get the aldehyde part of where you opened it, but what about the other carbon atom in the double bond (the one next to the CH3 branch), its a tertiary carbon atom, shouldn´t you get a ketone there?

Abby. Two B's.

__

Interesting. Well it's 2 am and my mind's numb. I'll look into it again tomorrow. Removing the answer for now to avoid confusion among the masses.

 

[zackle09]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_qp_42.pdf

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w10_ms_42.pdf

can someone please tell me how to do part 3 of question 3bi

 

[Young Stunner]

Can someone tell me how is this 1 & 2

 

[Faaiz Haque]

6:
I'd have picked D.

It's between A and D for me. I'll take D because you're getting rid of the double bond so 120 -> significantly lesser.

In A, and Alcohol will be formed but I'm not sure what the bond angles are. Should be lesser on one of the bonds due to the O lone pairs. Still I'd go for D because of the double bond -> single bond.
Edit: I just figured, it's the central carbon atom bond that we need to see not the O-H bond so A is out. A angles are all 109.5, the ones that matter.
17:
D

A- Happening.
B- NH3 -> NH4
C- (NH4)+ and (SO4)2-
D- You're just left with D. If you're unsure and want to waste more time, SO3 s = +6 final S = (-2+S-8)=0 so S= +6 so no redox.

22-
Free Radical Substitution
D

You're subtituting the H with Cl. Remember the very first reaction? FRSR of Alkane? That's what is happening. Don't let the acid part fool you it's the alkane part that's reacting here.

24-
D
Nucleophilic Substitution so both are nucleophiles

25-
B
A gives you propene
B gives you 2 bromo propane
(Remember, Halogenoalkanes, if theres a reaction that's not under ethanolic condition, it'll form an alcohol)
C gives you propene
D Again gives you propene (dehydration of the alcohol)


26:
Oxidation of a primary alcohol can give you Aldehydes and Carboxylic Acids.

27:
B
Ethanol reacting with Butanoic Acid so Ethyl Butanoate

28:
Will get back to you tomorrow.

29:
D

I can't explain this in words. Just draw it out and see where the double bonds would go. Adjacent Carbons
in this case. let me know if you dont get it and I'll draw them.

Thanks. I think i'll be able to do 29. And for 24, how do we know it's a Nucleophilic Reaction?

 

[Faaiz Haque]

Q20:

B- 2^3 = 6. If you know about statistics this is a bit easier. There are 2 possibilites around each double bond, either cis or trans. You have 3 double bonds in which this cis-trans combination can occur, hence it is 2 x 2 x 2 aka 2^3 = 6 (2 combinations for first double bond, 2 for second, 2 for third.)

Q38: B - 1 & 2 are correct. If you look at the beginning of the molecule from the left you can see that KMnO4 will break the double bond, and since the C attached to the double bond is a tertiary carbon atom (since its attached to 3 carbon atoms), it will yield, amongst other products, a ketone - in this case propanone. 2 is also correct because halogens can add across double bonds via the electrophilic addition mechanism.

Hope it helps.

Ty I understand 20
For 38 I get why #2 is correct but #1?. Can you explain it further? Why will KMnO4 break the double bond

 

[Abdel Moniem]

Thank you so much

c(iii) th gas in (ii) is CO2 and if you see total of gas formed is 40 but only 30 remains when the tottal gas is shaken with KOH(aq) se where did the 10 go it reacted with Koh(aq) and the gas is CO2 so 10cm^3
iv) they gave 50 O2 whcih gave us total of 40 cm^3 of gases of which 10 is CO2 so we have 30 left This 30 is O2 which is left from 50 so 50-30=20cm^3 got it?
d) the ms explains it quite well though but here we will use the volume ratio from balancing in b the ration between A used is 10 and CO2 produced is 10 so 1 cm^3 of A give 1 cm^3 of Co2 x=1
and 10 cm^3 of A reacts with 20 cm^3 of O2 so 1cm^3 of A reacts 2 cm^3 of O2
x=1
2=x+(y/4)
2=1+(y/4)
y=4
CxHy will be CH4

 

[Abdel Moniem]

Cyanohydrins can be made from carbonyl compounds by generating CN–ions from HCN in the presence of a weak base.

In a similar reaction, –CH2CO2CH3 ions are generated from CH3CO2CH3 by strong bases. Which compound can be made from an aldehyde and CH3CO2CH3 in the presence of a strong base?
A CH3CH(OH)CO2CH3
B CH3CO2CH2CH(OH)CH3
C CH3CH2CH(OH)CH2CO2CH3
D (CH3)2C(OH)CH2CO2CH3

 

[Namehere]

Ok!

Ty I understand 20
For 38 I get why #2 is correct but #1?. Can you explain it further? Why will KMnO4 break the double bond

It´s something you have to know. Acidified KMnO4 is such a strong oxidising agent that it will oxidise and break the double bond to produce CO2 + H2O if the cleavage occurs at a primary carbon atom; an aldehyde if the cleavage occurs at a secondary carbon atom or a ketone if the cleavage occurs at a tertiary carbon atom. Do note however, acidified potassium dichromate is only used as an oxidising agent and will not break double bonds since it is not a strong enough oxidisng agent (just in case, for future questions...)

Hope you understand.

 

[AbbbbY]

Q20:

B- 2^3 = 6. If you know about statistics this is a bit easier. There are 2 possibilites around each double bond, either cis or trans. You have 3 double bonds in which this cis-trans combination can occur, hence it is 2 x 2 x 2 aka 2^3 = 6 (2 combinations for first double bond, 2 for second, 2 for third.)

Ooo I missed this question.

Right method, wrong answer. 2^3 = 8, not 6

The answer should be C.


___


For 28, the answer is C. Three DNPH molecules will be needed to completely react.

http://i.imgur.com/vSkqtma.jpg
I messed it up a bit yesterday and forgot to draw the aldehyde leading to carboxylic acid.

This pic should explain the cleavage. If it's still unclear, let me know.

 

[Namehere]

Ooo I missed this question.

Right method, wrong answer. 2^3 = 8, not 6

The answer should be C.


___


For 28, the answer is C. Three DNPH molecules will be needed to completely react.

http://i.imgur.com/vSkqtma.jpg
I messed it up a bit yesterday and forgot to draw the aldehyde leading to carboxylic acid.

This pic should explain the cleavage. If it's still unclear, let me know.

Oh... 2^3 is indeed 8... my goodness i even said 2 x 2 x 2 = 6 :S This is what you get for not knowing how to multiply properly...

Q28 - I totally agree with your structure now! Before you missed the aldehyde turning into a carboxylic acid and the oxidation of the C at the double bond with the CH3 side chain.

Good work

 

[Muskan Achhpilia]

I have a doubt in the following questions of mcq can someone please help me...




Thank you so much!!

 

[Namehere]

I have a doubt in the following questions of mcq can someone please help me...
View attachment 38326

View attachment 38327

Thank you so much!!

Q35 - Strontium ions are a bit smaller than calcium ions due to the bigger nuclear charge, so I guess if you help yourself with the Data Booklet and search for ionic radius and stuff you can see they are about the same size and conclude 1 is correct.

Q31 - STATEMENT 3 is correct, nuclear charge does increase down the group (because you obviously have more protons...), but this does not answer why the bond length increases down the group. Actually, statement number 3 CONTRIBUTES to the bond length decreasing down the group, since it attracts those electrons closer to the nucleus, reducing the bond length. Statement 1 is correct - Actually CIE is pretty naught putting this stuff up since for a proper explanation you need a bit of knowledge of Molecular Orbital (MO) Theory. In a nutshell, orbital overlap decreases down the group because the bond stength decreases down the group. If you want a better explanation read on Band Gaps. Band Gaps decrease down the group, meaning the bond strength decreases down the group too, which reflects the decreasing amount of orbital overlap. If you think about it, you can follow this sequence: longer bond length -> bond strength decreases -> (because now you know a bit of band gaps), if bond strength decreases, band gaps decrease -> orbital overlap decreases.

Would you mind if I ask which year that paper is from? (Q31)

Hope it helps.

 

[halloffame]

Thanks

 

[Muskan Achhpilia]

Q35 - Strontium ions are a bit smaller than calcium ions due to the bigger nuclear charge, so I guess if you help yourself with the Data Booklet and search for ionic radius and stuff you can see they are about the same size and conclude 1 is correct.

Q31 - STATEMENT 3 is correct, nuclear charge does increase down the group (because you obviously have more protons...), but this does not answer why the bond length increases down the group. Actually, statement number 3 CONTRIBUTES to the bond length decreasing down the group, since it attracts those electrons closer to the nucleus, reducing the bond length. Statement 1 is correct - Actually CIE is pretty naught putting this stuff up since for a proper explanation you need a bit of knowledge of Molecular Orbital (MO) Theory. In a nutshell, orbital overlap decreases down the group because the bond stength decreases down the group. If you want a better explanation read on Band Gaps. Band Gaps decrease down the group, meaning the bond strength decreases down the group too, which reflects the decreasing amount of orbital overlap. If you think about it, you can follow this sequence: longer bond length -> bond strength decreases -> (because now you know a bit of band gaps), if bond strength decreases, band gaps decrease -> orbital overlap decreases.

Would you mind if I ask which year that paper is from? (Q31)

Hope it helps.

Thank you so much for the time and effort!

The question is from paper 1 summer 2006,
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s06_qp_1.pdf

 

[Muskan Achhpilia]

Can anyone please explain the following mcq questions...



For #12 I thought the answer should be C since Al2Cl6 is covalent while AlCl3 is ionic but it is B

Thank you!

 

[Namehere]

Can anyone please explain the following mcq questions...

View attachment 38357 View attachment 38358

For #12 I thought the answer should be C since Al2Cl6 is covalent while AlCl3 is ionic but it is B

Thank you!

Q13 - A is wrong because there is no hydrogen directly bonded to a very electronegative atom ie. Oxygen, Nitrogen or Fluorine.
B is wrong because there is only one molecule... dimeric would be like Al2Cl6 - two AlCl3 molecules bonding to each other, in this case via dative bonding.
C is wrong because the Al atom does have a complete octect of electrons - Al shares one electron with each of the three hydrogens, giving six electrons around the Al atom (2 + 2 + 2 ), the (CH3)3N part is bonded to the Al via dative bonding, since AlCl3 has one lone pair (2 electrons) which can be used for dative bonding. Adding together you get 6 (from the H part) + 2 (from the AlCl3 lone pair) = 8 electrons.
D is right because this molecule is like the CH4 molecule. No lone pairs and four molecules/group of molecules attached to the central atom. Bond angle of 109.5, with a tetrahedral arrangement.

Q12 - Well for this question, clearly A and D are wrong. A is completely ionic and SiCl4 is completely covalent. For C you may have misinterpreted that because Al (a metal) is bonded to H (a non-metal), it is ionic. However, AlCl3 is a covalent molecule! If it wouldn´t it couldn´t form the Al2Cl6 molecule, since to form that, two covalent AlCl3 molecules have to bond to each other via dative bonds (like i explained for the above question). You do write the AlCl3 molecule and SiCl4 molecule with lines to depict that the atoms are covalently bonded. You do not, however, write lines with NaCl and MgCl2, because they are formed from losing and gaining electrons and hence becoming ions, not by sharing electrons like you do in covalent bonding. B, MgCl2, has some covalent character because the Mg2+ ion is small and highly charged and therefore has a high charge density, meaning it could like to attract electrons to form bonds with. If you see, AlCl3 is a covalent molecule, because even that Al is a metal, the Al3+ is so highly charged (3+) and so small, that its huge charge density can attract the electrons closely enough so that the it can form covalent bonds with.

Hope it helps... I think I should write a bit less - too mucho info

 

[Muskan Achhpilia]

Q13 - A is wrong because there is no hydrogen directly bonded to a very electronegative atom ie. Oxygen, Nitrogen or Fluorine.
B is wrong because there is only one molecule... dimeric would be like Al2Cl6 - two AlCl3 molecules bonding to each other, in this case via dative bonding.
C is wrong because the Al atom does have a complete octect of electrons - Al shares one electron with each of the three hydrogens, giving six electrons around the Al atom (2 + 2 + 2 ), the (CH3)3N part is bonded to the Al via dative bonding, since AlCl3 has one lone pair (2 electrons) which can be used for dative bonding. Adding together you get 6 (from the H part) + 2 (from the AlCl3 lone pair) = 8 electrons.
D is right because this molecule is like the CH4 molecule. No lone pairs and four molecules/group of molecules attached to the central atom. Bond angle of 109.5, with a tetrahedral arrangement.

Q12 - Well for this question, clearly A and D are wrong. A is completely ionic and SiCl4 is completely covalent. For C you may have misinterpreted that because Al (a metal) is bonded to H (a non-metal), it is ionic. However, AlCl3 is a covalent molecule! If it wouldn´t it couldn´t form the Al2Cl6 molecule, since to form that, two covalent AlCl3 molecules have to bond to each other via dative bonds (like i explained for the above question). You do write the AlCl3 molecule and SiCl4 molecule with lines to depict that the atoms are covalently bonded. You do not, however, write lines with NaCl and MgCl2, because they are formed from losing and gaining electrons and hence becoming ions, not by sharing electrons like you do in covalent bonding. B, MgCl2, has some covalent character because the Mg2+ ion is small and highly charged and therefore has a high charge density, meaning it could like to attract electrons to form bonds with. If you see, AlCl3 is a covalent molecule, because even that Al is a metal, the Al3+ is so highly charged (3+) and so small, that its huge charge density can attract the electrons closely enough so that the it can form covalent bonds with.

Hope it helps... I think I should write a bit less - too mucho info

Thank you so much, it is great that you elaborate so well so the concept seems simple!

Can you help me with the following questions too...

PVC has single bonds because the double bonds breaks in chloroethene so always the single sigma bonds are stronger right?Though answer is B

I thought answer should be simply be A, and besides even in data booklet atomic radii of Neon is not mentioned so how is the answer D?

I am asking many question to you but seems like you are genius at chemistry...

Thanks a tonne!