10. Organic chemistry

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14 thoughts on “10. Organic chemistry

  1. 9701/MJ 2014/41
    Q 4.c)i) why sodium alkoxide is not formed together with phenoxide in reaction of noradrenaline with dilute NaOH?

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    • Noradrenaline (sometimes called norepinephrine), image here = http://wp.me/a4TWiX-59

      Are you wondering if the secondary alcohol of the noradrenaline may undergo deprotonation in the same way as the OH’s on the ring, i.e. the phenolic OH’s to form some kind of alkoxide (aliphatic chain with an O- on it)? I guess you are. If so, then consider this… when the OH- from NaOH extracts the proton of an aliphatic alcohol, H2O and an alkoxide ion would be formed. The alkoxide is a stronger base. It would deprotonate the newly formed H2O giving NaOH again and the alcohol, so the NaOH isn’t basic enough to abstract the H of the aliphatic alcohol. OK, it does happen a tiny bit, there is a little bit of acid base equilibrium, but the equilibrium lies heavily towards the secondary alcohol and not the secondary alkoxide. to form an alkoxide from acid-base reaction you need a much stronger base, one which likes protons more than the alkoxide you want to make. You may wonder about sodium(metal), that removes the H from aliphatic alcohols – yes? True, but that reaction is actually redox reaction. The H in the alcohol is +(I) and forms H2 which has an oxidation state of (0).

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    • The oxygen in arene alcohols (i.e. phenols) has its electron density delocalizing ‘into’ the arene ring due to the occurrence of the orbitals on oxygen overlapping with the ‘p’ orbitls of the arene ring. Hence the e- density on the oxygen is not great enough to attack the carboxyl carbon of a carboxylic acid.

      In an aliphatic alcohol, the is no (significant)overlap of the orbitals in oxygen and its attached hydrocarbon, so no delocalision occurs, hence in this case, the oxygen has enough electron density to attack the carboxyl carbon of a carboxylic acid, allowing the formation of esters.

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  2. Enantiomers are distinct molecules (i.e. have their own existence/identity) which are not superimposable on their mirror images.

    A bit more info:
    Just like cis and trans molecules can be uniquely identified/labelled by applying a method to them, another method can also be applied to determine a specifically label to any enantiomer:

    a) Locate the chiral centre (if there are no chiral centres then you will not get enantiomers!) in the molecule. The chiral centre will be an atom in the molecule which can produce a non-superimposable mirror image molecule, i.e. produce a different enantiomer by a different 3D positioning of the groups at that chiral centre atom.

    b) Using the R and S convention (Not on the syllabus) at the chiral centre.

    This system produces unique names for each enantiomer e.g.
    (R)-2-bromobutane
    and (S)-2-bromobutane.

    [Note: some enantiomers have more than one chiral centre. In such cases you may get molecules like:
    2(R)-chloro-3(R)-bromobutane
    2(R)-chloro-3(S)-bromobutane
    2(S)-chloro-3(R)-bromobutane
    2(S)-chloro-3(S)-bromobutane
    4 distinct molecules.

    Other molecules (starting from two or more chiral centres) can produce “meso” forms (also not on the syllabus).

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    • 9701 (Chemistry
      October/November Chemistry Paper 2, 2008 CE Chemistry, Question 2, sees the “ketene” molecule arise.

      Many of these organic problems are answerable using a very simple approach. Just compare the starting material with the end product.

      CH2C=C=O (ketene) —> CH2COOH (carboxylic acid)
      You can see the reactant:
      a) the PI BOND of the C=C was destroyed, only a sigma bond remained.
      b) the carbon in the CH2 picked up a H atom.
      c) The carbon in the =C=O group had OH added to it.
      c) The elements of water have ADDED to the ketene.
      This is a hydration reaction.

      CH2=C=O + H2O –> CH3COOH

      It’s true you’ve never seen a ketene in theory, and it’s also true two functional groups sharing the same C mean the properties of each functional group are no longer the same, as the two groups interact with each other resulting in ‘new’ properties (esters, amides, carboxylic acids, contain a C=O but you don’t get a positive test result with 2,4-DNP)
      But because this previously unknown molecule is not on your syllabus, you are expected to treat them as INDIVIDUAL groups., i.e. assume the ketene has the properties of an alkene and it also has the properties of a carbonyl, so in this case you would assume if asked that a ketene would undergo reaction with 2,4-DNP.

      You can think of this reaction as the O in H2O being nucleophilic enough to attack of the d+ of the =C= carbon. Then the usual mechanistic factors come into it. Overall you may notice a Markovnikov type addition to the C=C has occured.

      Always fall back on your basic chemistry.

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    • At this level of knowledge requirement, just look for a C atom with 4 different groups attached. The chiral centre is at that point. We have a chiral carbon at that point.

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  3. How does a reaction of an organic molecule with sodium with an observation of hydrogen gas evolved lead to a conclusion that a OH group being present?

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  4. It so happens that the H in OH group, can be reduced by Na(s), while other groups at A-level are not reactive with Na(s).

    The OH could either be in an alcohol, phenol or carboxylic acid. Hence it’s a reasonable test for OH groups.

    Note: As carbon chain lengthens, reactivity with Na falls.

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    • Updated:
      There is no functional group in poly(propene).
      The name “poly(propene)” is a relatively new way of naming the substance. The ‘new’ naming system is designed to show the monomer that the polymer came from. It’s shown in the round brackets part of the name. In the case of poly(propene) that means the monomer was propene. Once its polymerised however, the C=C that was present in propene, has gone – as one would expect for addition polymerisation – and you’re left with just a very long hydrocarbon which is actually an alkane and as you know, alkanes have no functional group.
      The polymer poly(ethene) is also an alkane having no functional groups.
      The polymer poly(chloroethene) is a poly chloroalkane (or poly halogenoalkane if you prefer).

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