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Chemistry: Post your doubts here!

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i am really confused what is the difference btw nucleophili addition and nucleophilic substution? can someone explain it to me properly?
 
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Q30:
Mr(Ethanol) = 46
Mr(Ethanoic acid) = 60
Mr(Ester) = 88

n(Ethanol) = 30 / 46 = 0.652
n(Ethanoic acid) = 30 / 60 = 0.50

That means ethanoic acid is the limiting reactant, so ester's yield should be checked by ethanoic acid and NOT by ethanol
n(Ester) = 22 / 88 =0.25

yield = (0.25 / 0.50) * 100% = 50%
So, C is the answer!

Q39:
During bromination of propane (C3H8),
the free radical will be .C3H7
It can be of two types:
(i) CH3 - .CH - CH3
(ii) .CH2 - CH2 - CH3

In option 1, one radical of type(i) and another radical of type(ii) are combining, so this is correct
In option 2, two radicals of type(ii) are combining, so this is also correct
However, in option 3, one butyl radical is combining with one ethyl radical, so this cannot be formed in bromination of propane!

This is why B is the correct answer!
Thank you! May Allah bless you :)
 
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9701_s11_ms_21.pdf
can anyone help me for question4 a)i and ii) i dont get why the answer is nucleophilic addition???
http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s11_ms_21.pdf
i am really confused what is the difference btw nucleophili addition and nucleophilic substution? can someone explain it to me properly?

Q4a(i) The answer is nucleophilic addition. This is the only type of reaction(as far as our AS chemistry syllabus is concerned) that carbonyl compounds (aldehydes and ketones) always undergo. Since the compound is propanone which is a ketone it will undergo nucleophilic addition. It is nucleophilic because a nucleophile attacks the C=O bond. It is addition because it involves addition of an atom or group of atoms in the C=O bond with the breaking of the double bond to form a single C-O bond and another C-something bond(depends on the reactants or reagents used).

Nucleophilic substitution always occurs in reactions of halogenoalkanes. It is when a nucleophile(a donator of a pair of electrons) substitutes a halogen in a halogenoalkane. If you go through the reactions of halogenoalkanes you will see that this is common for every reaction. for example reaction of a halogenoalkane to form an alcohol( reaction with an alkali). the nucleophile which in this case is the OH- ion from the alkali comes in place of the halogen in other words substitutes the halogen and an alcohol is formed. Hence it undergoes nucleophilic substitution.

Hope this clears your confusion!
 
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Q4a(i) The answer is nucleophilic addition. This is the only type of reaction(as far as our AS chemistry syllabus is concerned) that carbonyl compounds (aldehydes and ketones) always undergo. Since the compound is propanone which is a ketone it will undergo nucleophilic addition. It is nucleophilic because a nucleophile attacks the C=O bond. It is addition because it involves addition of an atom or group of atoms in the C=O bond with the breaking of the double bond to form a single C-O bond and another C-something bond(depends on the reactants or reagents used).

Nucleophilic substitution always occurs in reactions of halogenoalkanes. It is when a nucleophile(a donator of a pair of electrons) substitutes a halogen in a halogenoalkane. If you go through the reactions of halogenoalkanes you will see that this is common for every reaction. for example reaction of a halogenoalkane to form an alcohol( reaction with an alkali). the nucleophile which in this case is the OH- ion from the alkali comes in place of the halogen in other words substitutes the halogen and an alcohol is formed. Hence it undergoes nucleophilic substitution.

Hope this clears your confusion!
omg!!! thank u soooooooo much :) :LOL::cool:
 
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AoA,
Q7:
1 is tetrahedral so 109 degrees
2 is trigonal planar so 120 degrees [Remember that pie bons are not considered while checking bond angles!]
3 is bent / v-shaped so 105 degrees

So, smallest first, it is 3 < 1 < 2
So, C is correct!

Q8:
Hess Law:
posl.png
Enthalpy change = 4(-394) + 5(-286) - (-2877) = - 129 kJ mol-1
So, the answer is B

Q10:
Kc = [CH3CO2C2H5] [H2O] / [C2H5OH] [CH3CO2H]
4 = (x)(x) / (1-x)(1-x)
4 = (x / 1 - x)^2
(x / 1 - x)^2 = 4
Taking square root
x / 1 - x = 2
x = 2 - 2x
3x = 2
x = 2/3
So, B is the answer!
:)
 
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my answer is coming +129. Im using formula delta H of products - reactants . It is coming (-2877)-(4*-394)+(5*-286) = 129. What did i do wrong
 
Messages
1,477
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my answer is coming +129. Im using formula delta H of products - reactants . It is coming (-2877)-(4*-394)+(5*-286) = 129. What did i do wrong
It is not always products - reactants,
it depends on the diagram and the direction of the arrows!
 
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Solubility of hydroxides of Group II elements INCREASE down the group. So, 2 is wrong. There is no metallic bonding in hydroxyapatite. So, 3 is wrong. Then 1 MUST be right. Answer is D.
 
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Can somebody help me with finding good reliable notes on NMR spectroscopy???
 
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