Results are out! Find what you need...fast. Get quick advice or join the chat
Hey there Sign in to join this conversationNew here? Join for free

TSR Chemistry Society

Announcements Posted on
Applying to Uni? Let Universities come to you. Click here to get your perfect place 20-10-2014
    • 4 followers
    Offline

    ReputationRep:
    (Original post by Kallisto)
    As far as I know the charge of a neutron is neutral, whereas the charge of protons is positive and the one of electrons is negative. In terms of attraction force and centrifugal force neutrons have no influence on electrons due to the charge. So I think that the changing of number has no effects on electrostatic attraction, as neutrons neither positive nor negative. That is to say that neutrons are not able to repel or to attract electrons, even if the number of neutrons is greater in an isotope. That's why electrons will not come closer to the nucleus in an isotope. From this perspective and my considerations before I'm sure that isotopes are not able to reduce the atomic radius.
    You are correct.
    • 4 followers
    Offline

    ReputationRep:
    IR on its own is not really that diagnostic. You can't tell all that much from only an IR spectrum.

    (Original post by Ari Ben Canaan)
    Well, in the first one, you have the nice shoulder peaks just next to the C-H stretches at 2900 and then there's the sharp absorption at 1600. I'm guessing this is a very long chain aldehyde ?
    I don't think the first one has a carbonyl. The strech at ~1450 is too low for that. I reckon it's just a hydrocarbon of some sort, note the similarity of the spectrum to that of nujol. You also can't really tell the length of the hydrocarbon chain with any reliability from IR.

    The second one I think is an alkene ? I'm looking at the absorption to the left of 3000 and the single peak at 1600 as proof of this. An aromatic should have multiple absorptions between 1500 and 2500.
    Yeah presence of C-H single bonds and at least one C-H double bond. Could be aromatic though, but I would expect an alkene. Would need NMR or something else to be sure.

    The third one has a nice broad peak indicating an alcohol. The sharp peak to the left might be a free OH group ?
    Yeh an alcohol.

    Fourth one..... I'm going to go all in and say this is benzaldehyde. The two small peaks to the left of 2900 are definitely due to C-H stretching of the CHO group. The 1700 peak is a C=O stretch. The low intensity peaks at around 3000 and the multiple absorption at 700-900 are definitely indicative of an aromatic ring.
    Be careful here. These streches can be caused by alkenes as well, and the fingerprint region is not very diagnostic. While you may be correct in this case, I think you need to be careful about trying to take too much information from an IR spectrum. However, the spectrum is very similar to that of benzaldehyde so you may well be correct.

    Fifth one fits the bill of a secondary amine as there is only one peak at 3400.
    You missed the presence of a carbonyl as well, though I don't think that it is part of an amide as the C=O strech is too high.

    The last one is definitely a Carboxylic acid; that massively broad peak centered at 3000 is the overalp of the OH and sp3 hybridised C-H stretch. Then there's the sharp peak at 1700 due to the C=O stretch of the COOH group.
    Agreed.

    Overall, you're generally correct, but just be careful as sometimes I think you're coming to conclusions more detailed than you should really be based solely on these spectra.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by illusionz)
    IR on its own is not really that diagnostic. You can't tell all that much from only an IR spectrum.



    I don't think the first one has a carbonyl. The strech at ~1450 is too low for that. I reckon it's just a hydrocarbon of some sort, note the similarity of the spectrum to that of nujol. You also can't really tell the length of the hydrocarbon chain with any reliability from IR.



    Yeah presence of C-H single bonds and at least one C-H double bond. Could be aromatic though, but I would expect an alkene. Would need NMR or something else to be sure.



    Yeh an alcohol.



    Be careful here. These streches can be caused by alkenes as well, and the fingerprint region is not very diagnostic. While you may be correct in this case, I think you need to be careful about trying to take too much information from an IR spectrum. However, the spectrum is very similar to that of benzaldehyde so you may well be correct.



    You missed the presence of a carbonyl as well, though I don't think that it is part of an amide as the C=O strech is too high.



    Agreed.

    Overall, you're generally correct, but just be careful as sometimes I think you're coming to conclusions more detailed than you should really be based solely on these spectra.
    Hi illusionz and Ari Ben Canaan, Thank you very much for your help indeed I really appreciated all the help you have given me

    Illusionz can i just have confirmation for Spectra number 1 and number 5 please ?
    Especially number 5 because I cant understand from the post weather it is an amine or amide ?
    And can you tell me what bonds might be present in this spectra ?
    Thank you very much Lads
    • 18 followers
    Online

    ReputationRep:
    (Original post by illusionz)
    You are correct.
    Fine. But there is another question about isotopes: Why it is possible that there are atoms which have more neutrons than protons in a nucleus? in other words: How isotopes come into being?
    • 4 followers
    Offline

    ReputationRep:
    (Original post by D.Bman)
    Hi illusionz and Ari Ben Canaan, Thank you very much for your help indeed I really appreciated all the help you have given me

    Illusionz can i just have confirmation for Spectra number 1 and number 5 please ?
    Especially number 5 because I cant understand from the post weather it is an amine or amide ?
    And can you tell me what bonds might be present in this spectra ?
    Thank you very much Lads
    I think spectrum 1 is a hydrocarbon without a C=C, O, N etc It is possible there is a halogen but you really can't tell. You would need more data to determine the structure of the compound, NMR and mass spec would be very useful.

    As for 5, I do not think it is an amide as an amide C=O is generally ~1680 wavenumbers, and in any case, almost certainly below 1700. The C=O in this case is above 1700 which gives me the impression the C=O and N-H are seperate functional groups and not part of an amide.
    • 0 followers
    Offline

    ReputationRep:
    (Original post by illusionz)
    I think spectrum 1 is a hydrocarbon without a C=C, O, N etc It is possible there is a halogen but you really can't tell. You would need more data to determine the structure of the compound, NMR and mass spec would be very useful.

    As for 5, I do not think it is an amide as an amide C=O is generally ~1680 wavenumbers, and in any case, almost certainly below 1700. The C=O in this case is above 1700 which gives me the impression the C=O and N-H are seperate functional groups and not part of an amide.
    Mate for number 1
    Can you tell me what would be reason for the hydrocarbon to not have the C=C and O, N having looked at the spectra ?
    Is it due to the peaks or something ?
    • 4 followers
    Offline

    ReputationRep:
    (Original post by D.Bman)
    Mate for number 1
    Can you tell me what would be reason for the hydrocarbon to not have the C=C and O, N having looked at the spectra ?
    Is it due to the peaks or something ?
    Because the spectrum doesn't have peaks in the regions these bonds cause peaks to appear.
    • 0 followers
    Offline

    ReputationRep:
    Hey Everyone
    Can someone help me with a chemistry question

    I need to draw just one mechanisms for

    Nitration of Benzene
    Ethane + Bromine
    Ethane + Chlorine

    Can someone help me draw these please

    Thank you everyone xx
    • 11 followers
    Offline

    ReputationRep:
    (Original post by KellyRoth)
    Hey Everyone
    Can someone help me with a chemistry question

    I need to draw just one mechanisms for

    Nitration of Benzene
    Ethane + Bromine
    Ethane + Chlorine

    Can someone help me draw these please

    Thank you everyone xx
    For the nitration of benzene you are essentially reacting H2SO4 and HNO3 together to produce a Nitronium ion.

    Think about the interaction between the two acids.... Can you see a way by which a good leaving group (HINT: water) could be produced on the HNO3 molecule ?
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Ari Ben Canaan)
    For the nitration of benzene you are essentially reacting H2SO4 and HNO3 together to produce a Nitronium ion.

    Think about the interaction between the two acids.... Can you see a way by which a good leaving group (HINT: water) could be produced ?
    Right, So im essentially reacting :
    H2SO4 + HNO3 ----> Nitronium Ion ( Can you tell me the formula for this )

    So after reacting H2SO4 + HNO3 would i expect to see water being produced ?
    Can you correct me if am going wrong, because chemistry is my weakest subject
    Also what bonding would i see in this mechanism ? xx
    • 11 followers
    Offline

    ReputationRep:
    (Original post by KellyRoth)
    Right, So im essentially reacting :
    H2SO4 + HNO3 ----> Nitronium Ion ( Can you tell me the formula for this )

    So after reacting H2SO4 + HNO3 would i expect to see water being produced ?
    Can you correct me if am going wrong, because chemistry is my weakest subject
    Also what bonding would i see in this mechanism ? xx
    Are you an A Level student ?

    I'm sure you are aware that the electrophile is NO2+ i.e. a nitronium ion.

    Essentially, what happens is that the Nitric acid takes a proton from the Sulphuric acid to produce a good leaving group (water).

    What is left is the NO2+ ion.

    What follows is your usual electrophilic aromatic substitution reaction.

    Here, have a look at this :

    Click image for larger version. 

Name:	mechanism.png 
Views:	71 
Size:	38.8 KB 
ID:	190231

    EDIT : The reaction equation is
    HNO3 + 2H2SO4 ---> 2HSO4- + NO2+ + H3O+
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Ari Ben Canaan)
    Are you an A Level student ?

    I'm sure you are aware that the electrophile is NO2+ i.e. a nitronium ion.

    Essentially, what happens is that the Nitric acid takes a proton from the Sulphuric acid to produce a good leaving group (water).

    What is left is the NO2+ ion.

    What follows is your usual electrophilic aromatic substitution reaction.

    Here, have a look at this :

    Click image for larger version. 

Name:	mechanism.png 
Views:	71 
Size:	38.8 KB 
ID:	190231

    EDIT : The reaction equation is
    HNO3 + 2H2SO4 ---> 2HSO4- + NO2+ + H3O+
    Ah I see, it now makes sense
    I understand now that you've added a diagram in, and i can understand whats going on the reaction
    And yes i am doing a levels, its just i find chemistry a bit tricky

    One thing I am confused about is what type of reaction is it ?
    Is it electrophilic aromatic substitution like you have stated above ?
    And what type of bonding is going on in this mechanism ?
    x
    • 11 followers
    Offline

    ReputationRep:
    (Original post by KellyRoth)
    Ah I see, it now makes sense
    I understand now that you've added a diagram in, and i can understand whats going on the reaction
    And yes i am doing a levels, its just i find chemistry a bit tricky

    One thing I am confused about is what type of reaction is it ?
    Is it electrophilic aromatic substitution like you have stated above ?
    And what type of bonding is going on in this mechanism ?
    x
    The initial interaction between the two acids is a classic acid base reaction. Can you see how the Sulphuric acid donates a proton to the Nitric acid ? We then have a loss of water, which is secondary.

    What follows is an electrophilic aromatic substitution.

    Type of bonding ? What do you mean ?
    • 0 followers
    Offline

    ReputationRep:
    (Original post by Ari Ben Canaan)
    The initial interaction between the two acids is a classic acid base reaction. Can you see how the Sulphuric acid donates a proton to the Nitric acid ? We then have a loss of water, which is secondary.

    What follows is an electrophilic aromatic substitution.

    Type of bonding ? What do you mean ?
    Oh Right, So we first have an acid base reaction with a proton being donated to the nitric acid. Then a loss of water occours, which is secondary. Then electrophilic subsitution.
    I understand it now

    Ben Canaan, I was asked in the question to analyse the type of reaction in each case in relation to its bonding.
    This is the part where i am confused on, Can you help me please ?
    x
    • 18 followers
    Online

    ReputationRep:
    I'm back!

    I have discussed the fusion between Uranium and Deuterium in the physics society. During the discussion I have read that Uranium and Deuterium applies an energy bonding to become to Plutonium. Why energy come into being during the bonding between Uranium and Deuterium? I suppose that Uranium and Deuterium have to get a stimulated state to come to a bonding, as atoms keep the low-energy state by nature. And when atoms attend a bonding, they are stimulated. That is my supposition.
    • 0 followers
    Offline

    ReputationRep:
    Name: Aleisha Cullen
    Hobbies : singing, sport listening to music
    studying: AS chemistry, RS, psyschology and English literature
    live: westmiddlands
    why i like chemo: its logical and interesting
    • 0 followers
    Offline

    ReputationRep:
    please help me with this question !!!

    Given that
    p(g) + 3Cl(g) = PCl3 (g) energy change = -983kjmol-1
    P(s) + 3/2Cl2(g) = PCl3(g) energy change = -305kjmol-1
    p(s) = p(g) energy change = +314kjmol-1

    Calculate the following bond energies:
    a) P-Cl in PCl3

    b) Cl-Cl in Cl2

    I really have no clue on how to do this so please help :s
    • 11 followers
    Offline

    ReputationRep:
    (Original post by Aleisha1)
    please help me with this question !!!

    Given that
    p(g) + 3Cl(g) = PCl3 (g) energy change = -983kjmol-1
    P(s) + 3/2Cl2(g) = PCl3(g) energy change = -305kjmol-1
    p(s) = p(g) energy change = +314kjmol-1

    Calculate the following bond energies:
    a) P-Cl in PCl3

    b) Cl-Cl in Cl2

    I really have no clue on how to do this so please help :s
    Hint : Enthalpy change of reaction = Energy used to Break bonds - Energy used to form bonds.
    • 11 followers
    Offline

    ReputationRep:
    (Original post by Ari Ben Canaan)
    Hint : Enthalpy change of reaction = Energy used to Break bonds - Energy used to form bonds.
    I dont understand why enthalpy change = bonds broken - bonds formed


    Posted from TSR Mobile
    • 1 follower
    Online

    ReputationRep:
    hi could anyone give me some advice? Basically I am considering being a pharmacist but I don't know that much about it. What different things can a pharmacist do?

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: October 18, 2014
New on TSR

What is sixth form like?

Share your story!

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.