Chemistry: Post your doubts here!

[nehaoscar]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_13.pdf

Q29 ( ans is B)
30 (C) nd 20 (C) pleasee

Q20:

The green line shows the main carbon chain. These are the 5 isomers

 

[nehaoscar]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_13.pdf

Q29 ( ans is B)
30 (C) nd 20 (C) pleasee

O29 : X is wrong as if you chose any one carbon atom and draw a 3D structure around it, you can see that all the 4 different substituents are in different planes. So since the C has another C as one of the 4 substituents, they are in different planes.
Y is wrong because for cis-trans, you need 2 same substituents on the 2 carbons, in this case there are not.

 

[qwertypoiu]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_13.pdf

Q29 ( ans is B)
30 (C) nd 20 (C) pleasee

29 -All the carbon atoms bonded in the ring have tetrahedral structure, and are sp3 hybridised. this is because every carbon is attached to four groups. So like when you draw optical isomers, in fact, every carbon with 4 substituents have a 3 dimensional tetrahedral structure. Therefore, they cannot be on the same plane. Also, there is no Cis trans isomerism for a cycloalkene made up of less than 8 carbon atoms, since the ring constricts bond rotation. So neither students are correct.

30: the compound already given is:
CH3CH2CH2CO2CH3
Now just play around:

1. CH3CH2CH2CO2CH3
   
2. CH3CH(CH3)CO2CH3
3. CH3CH2CO2CH2CH3
4. CH3CO2CH2CH2CH3
5. CH3CO2CH(CH3)CH3
6. HCO2CH2CH2CH2CH3
7. HCO2CH(CH3)CH2CH3
8. HCO2CH2CH(CH3)CH3
9. HCO2C(CH3)3

20. Again just play around

1. CH3CH2CH2CH2CHO
2. CH3CH2C*H(CH3)CHO
3. OPTICAL ISOMER OF ABOVE
4. CH3CH(CH3)CH2CHO
5. C(CH3)3CHO

 

[nehaoscar]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_13.pdf

Q29 ( ans is B)
30 (C) nd 20 (C) pleasee

Q30:



These are the 9 isomers

 

[princess Anu]

Thanks a lot guys.. ^_^

 

[princess Anu]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_13.pdf
Q19: Why is B incorrect?

 

[nehaoscar]

Why A?

 

[nehaoscar]

How to do this?
Answer is B

 

[nehaoscar]

HOW TO DO THIS ONE?

 

[nehaoscar]

Why A? why not the others?

 

[nehaoscar]

Answer is D but why not C?

 

[nehaoscar]

???

 

[Mathemagical]

Thanks i do understand 27 and 29 now but can you please explain 22 a bit more? thanks again

The keyword in Question 22 is "followed by". Take Option B, for example. After polymerising it, the alkene becomes saturated and therefore cannot undergo a reaction with cold, dilute KMnO4. Similarly, alkanes, which are also saturated, cannot undergo that reaction.

 

[Mathemagical]

View attachment 54229

Answer is D but why not C?

The first reaction in which the intermediate is formed is nucleophilic substitution. For nucleophilic substitution, either sodium cyanide or potassium cyanide in ethanol can be used.

Hydrogen cyanide is used in nucleophilic addition reactions, for example with aldehydes or ketones.

 

[Metanoia]

Q20:
View attachment 54222
The green line shows the main carbon chain. These are the 5 isomers

Your bottom two isomers are identical.

Instead, find the chiral carbon in the 2nd isomer (top row), indicating that there is an extra optical isomer.

The question has hinted to look for optical isomers, so likely there will be at least one.

 

[Metanoia]

View attachment 54225
Why A?

Compared to the others, CO2 has the strongest intermolecular forces of attraction.

An ideal gas is supposed to have negligible intermolecular forces of attraction.

 

[Mathemagical]

View attachment 54228
Why A? why not the others?

The equilibrium constant is only affected by temperature, and not by anything else.

 

[Mathemagical]

View attachment 54227
HOW TO DO THIS ONE?

The answer is A, sodium chloride.
Upon addition of dilute aqueous ammonia, the precipitate dissolves, so chloride ions are present.

 

[Studydayandnight]

how do we solve this one? :/

Correct answer is C.

 

[Mathemagical]

http://maxpapers.com/wp-content/uploads/2012/11/9701_w14_qp_13.pdf
Q19: Why is B incorrect?

The melting points of SiO2 and CaCo3 are 1600°C and 825°C respectively. Such high temperatures cannot be achieved by using a Bunsen burner, and so both of them would not melt under a Bunsen flame.