5A Alkenes - Edexcel IAS Chemistry (Unit 1)

00:34:26
https://www.youtube.com/watch?v=5GUi5Up-ZPc

Resumo

TLDRStu video esamina i dettagli d'alcheni cum'è parte di u cursu di chimica IES da Edexcel. Si discute di a formula generale di l'alcheni, confrontanduli cù ciclu alcani è identificendu a so insaturazione per mezu di u ligame doppiu carboniu-carboniu. Inoltre, si tratta l'isomerismu geometricu cù i sistemi di nomenclatura E/Z è cis/trans. Si esploranu e reazioni d'alcheni, cumpresu i test qualitativi per ligami doppi cù bromu o acqua bromata, è i mecanismi di l'addizione elettrofilica di bromu è bromuro d'idrogenu. Si enfatiza nant'è e proprietà reattive d'alcheni, sopratuttu cuncernendu i ligami pi chì sò più suscettibili à reagisce. Infine, hè discussu l'isomerismu stereochimicu limitatu da a rotazione à mezu à i ligami doppi.

Conclusões

  • 🧪 Gli alcheni sò idrocarburi insaturati cù un ligame doppiu carboniu-carboniu.
  • 🔍 Si tratta di isomerismu geometricu cù i sistemi cis/trans è E/Z.
  • 💡 I ligami pi, essendu esposti, rendenu l'alcheni più reattivi.
  • 🧊 Test qualitativi per ligami doppi usanu acqua bromata.
  • ⚙️ Meccanismi di addizione elettrofilica per bromu è bromuro d'idrogenu.
  • 🔗 Rotazione limitata in ligami doppi porta a isomerismu stereochimicu.
  • 🌿 L'alcheni ponu esse furmati per mezu di crackazione di oliu crudu.
  • 🔥 Reazzioni d'addizione cù alcheni formanu prudutti saturati.
  • 📊 Stabilità di carbocationi determina u pruduttu maestru in reazioni.
  • 🔬 Applicazioni pratiche di nomi sistematici in chimica organica.
  • 🎓 Aspetti d'esami in meccanismi di reazioni Alcheniche.

Linha do tempo

  • 00:00:00 - 00:05:00

    吾嘎里视频喀谈论到诸如烯烃的基本化学性质，包括一般公式CₙH₂ₙ以及环烷烃的比较。还有，吾嘎里喀探讨了碳-碳双键的细节, 以及如何使用EZ或顺反系统命名几何异构体。此外，它解释了如何进行烯烃的反应，尤其是配以溴或溴水进行的双键定性试验，及溴和氢溴酸的亲电加成机制。

  • 00:05:00 - 00:10:00

    烯烃的高反应性来自碳-碳双键，特别是π键。吾嘎里解释了σ键和π键的区别，并且π键的电子离碳原子较远，使其更易参与反应。吾嘎里还重新提及在前一主题中曾提到的不同类型的异构现象，特别是几何异构。几何异构体由于π键的存在会有旋转限制。

  • 00:10:00 - 00:15:00

    然后, 吾嘎里继续探讨几何异构体如何因其在分子中的地位而不同, 并解释了如何使用EZ命名系统来确定较优先的基团。吾嘎里提供了几个例子说明如何分辨顺-反异构体与EZ异构体，利用原子序数来标定优先级，进而辨别其在双键上的位置。

  • 00:15:00 - 00:20:00

    吾嘎里描述了烯烃加成反应的多种实例，其中包括氢化，卤化，水合及氧化反应的实例。特别是在氧化反应中，醇与氧化剂及水反应形成具有两个羟基的二醇。氧化剂通常为酸化高锰酸钾，引起颜色由紫转无色的改变。

  • 00:20:00 - 00:25:00

    加入了电解亲电加成机制，介绍了卤素的亲电加成及氢卤酸的机制。吾嘎里使用曲箭表示电子对的移动，以及解释如何形成羰基正离子中间体。吾嘎里特别提到烯烃如何因有π键形成羰基正离子中间体并与卤负离子反应。

  • 00:25:00 - 00:34:26

    最后，吾嘎里总结说，必须熟悉主观性烯烃的不同可能产物，强调次级羰基正离子比初级羰基正离子更稳定，因此作为主要产物。且演示了在考试中可能出现的问题，例如描述反应步骤及绘制分子结构。并解释了为什么像布罗莫一和二溴化物烃会有不同的产物经济性。

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Mapa mental

Mind Map

Perguntas frequentes

  • Chì sò l'alcheni?

    Sì, l'alcheni sò idrocarburi insaturati per via di u ligame doppiu carboniu-carboniu.

  • Perchè l'alcheni sò più reattivi chè l'alcani?

    I ligami sigma sò più forti di i ligami pi, chì rendenu l'alcheni più reattivi per via di i ligami pi esposti.

  • Chì hè u test chimicu per l'alcheni cù l'acqua bromata?

    A reazione include l'aghjunzione di bromu attraversu u ligame doppiu, risultendu in a decolorazione di l'acqua bromata.

  • Cumu si manifesta l'isomerismu geometricu in l'alcheni?

    A rotazione hè limitata intornu à u ligame doppiu carboniu-carboniu, risultatu in differente pusizioni per i gruppi attaccati.

  • Cumu si determina a stabilità di un carbocatione?

    I carbocationi più stabili sò quelli chì sò cunnessi à più gruppi alkili.

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Legendas
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Rolagem automática:
  • 00:00:00
    [Music]
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    and that's penny we're gonna be looking
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    at topic 5e which has the alkenes and
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    this is part of the IES chemistry course
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    from in Excel so we'll be looking at the
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    general formula for the alkenes and then
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    comparing them to the cyclo alkanes
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    looking at their unsaturation and here
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    we actually go into the detail of a
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    carbon-carbon double bond we'll be
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    looking at geometric isomerism and how
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    we use the easier naming system as well
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    as the strands naming system to identify
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    these isomers we'll look at the
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    reactions of alkenes and particularly
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    qualitative tests for a double bond
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    using bromine or bromine water and also
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    the mechanisms for the electrophilic
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    addition of bromine and the
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    electrophilic addition of hydrogen
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    bromide so alkenes are unsaturated
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    hydrocarbons and this is because they
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    contain a carbon to carbon double bond
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    we generally form them and cracking
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    reactions so we underwent our fractional
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    distillation of our crude oil and we
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    have formed our substance that we then
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    undergo cracking in order to make these
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    shorter more useful molecules and we
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    tend to form alkenes and all our kings
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    will share the same general formula of
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    cnh2n and due to the presence of this
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    double bond our kings
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    tend to be much more reactive than
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    alkenes and we'll look at the reasons
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    for that in just a moment in terms of
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    the nomenclature we follow the same
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    rules as what we did for the alkenes but
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    this time rule number two we want to
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    number so that the double bond is on the
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    lowest possible number so we identify
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    the longest continuous chain number the
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    carbons and then identify the number of
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    possession and type of branches and then
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    we name it so let's look at two examples
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    or we've got CH 3 CH 2 CH with our
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    branch CH 2 c8
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    ch2 so we can see that my longest
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    continuous chain as the center chin and
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    looking right away we can see that the
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    double bond is going to be on the lowest
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    number every number from right to left
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    so we're gonna go one two three four
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    five and six so I know if I have sex
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    carbons
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    I have hex now I have to specify where
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    the double bond does and the double bond
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    is between carbons number one and two
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    and I use the lure of the two numbers so
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    this becomes hex one in because I can
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    also have the double bonds in different
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    places I have to specify it here I also
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    have that side chin which has a methyl
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    and that is on carbon number four so
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    this is for me file hex one in if I look
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    at the second example again my longest
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    continuous chain is on the center and
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    the find number from right to left
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    I'm gonna get the lowest possible number
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    so I have one two three and four my
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    carbon-carbon double bond is between
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    carbon number one and two again and this
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    as a fortune so this is Butte one in and
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    I have two methyl groups so I have
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    methyl and I have died giving me 3 3 so
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    this is diet and we fell Butte one so
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    I'm following the same rules here that I
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    did for the alkanes no major defens the
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    only thing is making sure that the
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    double bond is on the lowest possible
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    carbon now when we're looking at carbon
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    two carbon double bonds what we need to
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    understand is that two bonds are not the
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    same one of them is a segment and one of
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    them as a PI bond
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    so remember Sigma bonds as when the PI
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    orbitals overlap in to end that's our
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    Sigma and then PI is when the overlap
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    say to side like that so
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    segment we get one area of overlap and
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    our pie we get two idiots of overlap and
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    that we get both of these overlaps
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    happening to form our double bond so we
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    get our axial overlap otherwise known as
  • 00:04:42
    our end to end and then that one extra
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    electron and the P orbitals
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    give us our site to site overlap and
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    it's important that we're able to draw
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    both of those out because that can come
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    up in an exam so we can see here we've
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    already made our segment bond and then
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    we have these PI orbitals sorry P
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    orbitals that are left over and B then
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    formed s PI bond now what we need to
  • 00:05:07
    understand is that the electrons and a
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    PI bond are then further away from the
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    carbon atoms and comparison to a segment
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    so in a segment bond the electrons are
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    quite close to the carbon atoms so
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    they're held quite tightly I'm more
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    attracted with as in a PI bond they're a
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    little bit further away because they're
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    that bit further away and they're less
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    attracted then they are more likely to
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    undergo reactions and to take part and
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    reactions with other molecules therefore
  • 00:05:38
    our kings because they have this pi bond
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    are more reactive than our kids because
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    alkenes only contain segments now in
  • 00:05:50
    topic 4 we talked about different types
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    of isomerism and we looked at structural
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    isomerism so we were looking at chain
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    and possession eyes organism and we can
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    also have our functional grip as well
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    but we have a went into too much detail
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    about that just yet because we haven't
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    looked at too many different functional
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    groups
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    now unless topic we're going to look at
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    something known as steady or isomer and
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    in particular we're focusing on
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    geometric isomers optical isomers is
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    something that you will cover and topic
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    15 and a two so we're gonna focus on
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    geometric so you've looked at structural
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    eyes on that as and before now we're
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    looking at stereoisomerism
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    now geometric isomers def
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    each other in terms of structure best
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    way because of where their groups are
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    positioned on the molecule so rather
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    than it being just simply moving a chain
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    from carbon number two to carbon number
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    three we're looking at the molecule as a
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    whole and it's because they are going to
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    be on different sites of the carbon two
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    carbon double bond so geometric isomers
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    can only exist and alkenes they can't
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    exist in alkanes and that is because
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    this carbon two carbon double bond leads
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    to restricted rotation if we have only
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    second my bonds we can have a rotation
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    around our carbon atoms and they can
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    move and they can be twisted when we
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    have carbon two carbon double bond that
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    has the segment and PI bonds this
  • 00:07:28
    rotation is restricted so we can't get
  • 00:07:31
    this twisting of the molecule so we can
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    only get these groups attached to the
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    carbon two carbon double bond and one of
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    two possessions and we can use two
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    different naming conventions we can use
  • 00:07:43
    sess and trans or e ends it and we're
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    going to look at both of these so let's
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    start off by looking at this s and the
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    trans so n last case we're looking at
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    butte tui
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    so butte two even has the carbon two
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    carbon double bond and the center so if
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    i draw it just very easily stretching
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    you can see we have something that looks
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    like that okay now when we're comparing
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    the two structures that we have down
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    here on the left-hand structure our ch3
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    groups are on opposite sides of the
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    double bond okay so we're talking about
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    top and bottom not left and right so we
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    want to put plane of symmetry through
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    the center and they're on opposite sides
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    and we call this trans view too in when
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    they are on opposite sides when we do
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    the same thing and we put this Lane of
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    symmetry and we compare for the second
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    structure the CH threes are on the same
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    side of the double bond and we call this
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    sess Butte to you so if they are on
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    opposite sides its trans and if they are
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    on the same side assess now the problem
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    with us is this particular naming system
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    only works for very specific compounds
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    ie when we have two groups attached as
  • 00:09:12
    soon as you start to have different
  • 00:09:14
    numbers of groups three or four groups
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    attached to your carbons then we run
  • 00:09:19
    into some problems how do you know what
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    ones you want to compare so this is
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    where we use something known as the easy
  • 00:09:25
    naming system so an example of a
  • 00:09:29
    molecule that would use the easy naming
  • 00:09:31
    system as this one here so you can see
  • 00:09:33
    that we have a hydrogen coding a
  • 00:09:36
    floating and I'm throwing all attached
  • 00:09:39
    to the carbon to carbon double bond how
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    do we know what ones to compare well
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    this is what the easy system is going to
  • 00:09:46
    show us the first thing that we do is we
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    work out the name that's going to be
  • 00:09:52
    used for both of the isomers and we just
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    use our normal nomenclature rows so in
  • 00:09:56
    this case our name as one bromo one
  • 00:09:59
    coral to flirt o ething and we used to
  • 00:10:03
    use our normal finding the longest
  • 00:10:05
    continuous chain identifying our chains
  • 00:10:08
    and making sure that our atoms that are
  • 00:10:11
    attached are put an alphabetical order
  • 00:10:13
    then we use something known as priority
  • 00:10:16
    rules to decide which of the two atoms
  • 00:10:19
    on the left-hand side of the double bond
  • 00:10:21
    is the higher priority so in steps 2 & 3
  • 00:10:25
    we are looking at left and right in
  • 00:10:28
    terms of the carbon so we compare the
  • 00:10:31
    hydrogen and the floating and we look at
  • 00:10:33
    which one has the higher atomic number
  • 00:10:36
    and that's the one that takes priority
  • 00:10:38
    so for hydrogen it has an atomic number
  • 00:10:41
    of 1 whereas floating has an atomic
  • 00:10:44
    number of 9 so floating is the higher
  • 00:10:47
    priority then we look at the right-hand
  • 00:10:51
    side and we compare coating with atomic
  • 00:10:54
    number 17 and bromine with atomic number
  • 00:10:57
    35 and the bromine has the higher
  • 00:11:00
    priority so I'm just gonna mark the same
  • 00:11:03
    on the other then for step four we're
  • 00:11:06
    looking at
  • 00:11:07
    top and bottom so I use that line of
  • 00:11:11
    symmetry through the center and then I
  • 00:11:14
    compare my two priority groups F both of
  • 00:11:18
    my priority groups are on the same side
  • 00:11:21
    either both above or both the low then
  • 00:11:24
    this is the Z isomer so the one on the
  • 00:11:27
    left as the Zed and if the groups are
  • 00:11:30
    one above and one below then this as the
  • 00:11:34
    e isomer so it's similar to this s and
  • 00:11:38
    the trans in terms of looking at which
  • 00:11:41
    one is above and below so E is similar
  • 00:11:45
    to trans and Z a similar to cess but the
  • 00:11:48
    easy naming system uses these priority
  • 00:11:51
    rules and looks at the atomic number and
  • 00:11:53
    then applies look the above or below the
  • 00:11:57
    double bond so I would strongly suggest
  • 00:12:00
    that you practice this because it does
  • 00:12:02
    come up very commonly and multiple
  • 00:12:04
    choice questions now when we have this
  • 00:12:10
    carbon-carbon double bond made up of the
  • 00:12:13
    segments and the PI bonds what we need
  • 00:12:14
    to understand is that a sigma bond is
  • 00:12:17
    stronger than a pi bond but it is not
  • 00:12:20
    twice as strong and most alkenes involve
  • 00:12:24
    the double bond becoming a single bond
  • 00:12:26
    so what this actually means is that my
  • 00:12:28
    segment bond remains unchanged and the
  • 00:12:32
    electrons that are and the PI bond are
  • 00:12:34
    used to form the new bonds so the
  • 00:12:37
    segment bond is left and the PI bond is
  • 00:12:41
    broken and this will then form a
  • 00:12:43
    saturated product that only contains
  • 00:12:46
    segments now you've seen an example of
  • 00:12:49
    an addition reaction back at GCSE so the
  • 00:12:54
    chemical test for an alkene has the
  • 00:12:56
    bromine water test and thus as an
  • 00:12:59
    addition reaction and as I said the
  • 00:13:01
    student beam yure you did learn this
  • 00:13:03
    back in a GCSE but what we now know as
  • 00:13:06
    we can go until a bit more detail in
  • 00:13:08
    terms of the actual bond and the PI bond
  • 00:13:11
    and the alkyne breaks when the bromine
  • 00:13:15
    is added across the double bond and then
  • 00:13:17
    we form the colourless product of
  • 00:13:21
    gramo even causing this decolorization
  • 00:13:24
    of the bromine water and that's the
  • 00:13:26
    chemical test so if we want to identify
  • 00:13:29
    the presence of an our kin we will
  • 00:13:32
    decolorize a bromine water and it will
  • 00:13:35
    be because we're breaking this pi bond
  • 00:13:37
    to make a saturated compound and you can
  • 00:13:39
    see that and then picture at the bottom
  • 00:13:45
    now we have four key examples of
  • 00:13:48
    addition reactions the first is
  • 00:13:50
    hydrogenation which is the addition of
  • 00:13:53
    hydrogen in order to form an alkene and
  • 00:13:55
    this requires a nickel catalyst we can
  • 00:14:02
    also have halogenation which is the
  • 00:14:04
    addition of a halogen in order to form
  • 00:14:07
    something known as a dialogical alkene
  • 00:14:09
    meaning I have two halogens on an alkene
  • 00:14:12
    and we'll talk a little bit more about
  • 00:14:14
    that in just a few minutes we can also
  • 00:14:17
    have a hydration which is the addition
  • 00:14:20
    of water to form an alcohol and this
  • 00:14:22
    requires a phosphoric acid catalyst or
  • 00:14:26
    HTTP or four and we can form something
  • 00:14:29
    like ethanol and we can also add a
  • 00:14:32
    hydrogen halide to form a halogen or
  • 00:14:36
    alkyne meaning an alkane with only one
  • 00:14:39
    halogen attached this time we can also
  • 00:14:45
    have a reaction known as an oxidation of
  • 00:14:48
    an alkene and this is going to form
  • 00:14:50
    something known as a diol and what
  • 00:14:54
    actually happens is we get both an
  • 00:14:56
    addition reaction because we get the
  • 00:14:58
    breaking of a double bond and we get an
  • 00:15:00
    oxidation with oxidation being adding
  • 00:15:05
    oxygen and the dial is a compound that
  • 00:15:09
    contains two Oh H grips and you'll come
  • 00:15:13
    back to looking at which groups when you
  • 00:15:14
    get to topic 10 but we will look at it
  • 00:15:17
    just briefly here in terms of when it
  • 00:15:20
    reacts
  • 00:15:20
    from an alkene so we use an oxidizing
  • 00:15:24
    agent in order to supply this oxygen and
  • 00:15:27
    is generally acidified potassium
  • 00:15:29
    manganate to seven okay so we put the
  • 00:15:33
    potassium
  • 00:15:34
    staticky conditions usually with
  • 00:15:37
    sulfuric acid and the potassium
  • 00:15:39
    manganate provides an oxygen and the
  • 00:15:42
    water is going to pervade a hydrogen and
  • 00:15:45
    a second oxygen and it causes to which
  • 00:15:48
    groups to be added across the double
  • 00:15:51
    bond so we're breaking up the double
  • 00:15:54
    bond adding on some water or with the
  • 00:15:57
    potassium manganate and we get to all
  • 00:16:00
    age groups being added so we get a
  • 00:16:03
    reaction like this we have the ething
  • 00:16:08
    this is our oxidizing agent we just
  • 00:16:11
    simply write it as an all and square
  • 00:16:13
    brackets plus water and then we make
  • 00:16:17
    Ethan one to dial and this actually
  • 00:16:21
    looks like that so we have 208 groups
  • 00:16:30
    being attached and we see a color change
  • 00:16:32
    from the potassium manganate from purple
  • 00:16:34
    to colorless and we can also use this as
  • 00:16:38
    another identifying test for alkenes
  • 00:16:40
    because our Kings will not undergo this
  • 00:16:42
    reaction only our Kings can undergo
  • 00:16:45
    oxidation to a dial now when we are
  • 00:16:50
    looking at addition reactions we need to
  • 00:16:52
    understand the mechanism we looked at a
  • 00:16:55
    mechanism when we looked at few radical
  • 00:16:57
    substitution and we looked at curly
  • 00:16:59
    arrows in this case we're going to be
  • 00:17:02
    using the curly arrows and much more
  • 00:17:04
    detail and remember the curly arrows
  • 00:17:07
    show the movement of the electron pair
  • 00:17:11
    and that's going from the bond to an
  • 00:17:16
    atom or it could be going from a lone
  • 00:17:20
    pair to form a bond those are the two
  • 00:17:25
    different types of Carly arrows that we
  • 00:17:27
    see so we're going to be looking at the
  • 00:17:29
    addition of hydrogen bromide to Ethan
  • 00:17:32
    but the same principles can apply to any
  • 00:17:35
    hydrogen halide to any alkyne and
  • 00:17:38
    because our kings have pi bonds they are
  • 00:17:42
    what we call an area of high electron
  • 00:17:45
    density meaning there are what
  • 00:17:47
    of electrons and not area and the
  • 00:17:50
    hydrogen bromide is a polar molecule so
  • 00:17:53
    as each BR and we get a delta positive
  • 00:17:57
    and a delta negative because my
  • 00:18:02
    electrons are closer to the bromine and
  • 00:18:04
    then what happens is less delta positive
  • 00:18:06
    hydrogen is going to be very strongly
  • 00:18:09
    attracted to this pi bond because it has
  • 00:18:13
    this area of high electron density so we
  • 00:18:18
    describe that Delta positive end of hbr
  • 00:18:22
    as as an electrophile so the
  • 00:18:27
    electrophile likes an area of negative
  • 00:18:30
    charge to end next case the PI bond and
  • 00:18:32
    then we're going to be using our curly
  • 00:18:35
    arrows to show the movement and the
  • 00:18:37
    curly arrows after however have to start
  • 00:18:39
    from a bond and go to an atom or go from
  • 00:18:43
    a lone pair and move to an atom and move
  • 00:18:46
    to form a bond and we'll see a lot but
  • 00:18:49
    more of that and detail as we go through
  • 00:18:51
    the a level course and we start to
  • 00:18:53
    become more used to using curly arrows
  • 00:18:57
    so let's look at the actual
  • 00:18:59
    electrophilic addition at this point so
  • 00:19:03
    the reaction begins with a heterolytic
  • 00:19:07
    fashion so this allows both of the
  • 00:19:10
    electrons to go to one of the atoms and
  • 00:19:13
    what we actually mean by that is we're
  • 00:19:15
    breaking up the hydrogen bromide and
  • 00:19:18
    both of the electrons are going to go to
  • 00:19:21
    the put all made I am rather than the
  • 00:19:24
    hydrogen so we get a curly arrow coming
  • 00:19:28
    from the electrons in the PI bond moving
  • 00:19:31
    into the hydrogen meaning we are forming
  • 00:19:34
    a bond between the electrons and this pi
  • 00:19:38
    bond and this hydrogen so the carbon is
  • 00:19:41
    going to be bonded to the hydrogen like
  • 00:19:46
    this
  • 00:19:47
    so this Carly Otto forms that bond we
  • 00:19:52
    also see that's curly arrow where we
  • 00:19:55
    this is the movement of the electrons so
  • 00:19:58
    this is the heterolytic
  • 00:20:01
    fashion and thus forms thus bromate iron
  • 00:20:06
    and the other substance that is formed
  • 00:20:10
    is known as a carbo-cation and a
  • 00:20:14
    carbo-cation is just a substance that
  • 00:20:16
    has a positive charge because anthea
  • 00:20:19
    here thus carbon as messing a bond so it
  • 00:20:26
    forms a positive charge and this is
  • 00:20:29
    known as the carbo cation intermediate
  • 00:20:30
    and then we have our bromide iron with a
  • 00:20:34
    lone pair so this is the first step the
  • 00:20:38
    second step now is that this lone pair
  • 00:20:40
    on the bromide is going to be attracted
  • 00:20:43
    to the area of positive charge on the
  • 00:20:46
    carbon cation so it will donate the
  • 00:20:50
    electrons and to the carbo cation so
  • 00:20:55
    we're going to see this movement here
  • 00:20:56
    again forming a bond and this time we're
  • 00:20:59
    forming this c BR bond in this case were
  • 00:21:04
    forming bromo Ethan so we had our
  • 00:21:08
    heterolytic fashion in order to form our
  • 00:21:13
    through my die and our carbo cation
  • 00:21:15
    showing our Carly arrows
  • 00:21:17
    note the bromide ion is attacking the
  • 00:21:20
    carbo cation and then we're going to
  • 00:21:22
    form this covalent bond in order to make
  • 00:21:25
    my bromo ethan so there are the two
  • 00:21:30
    steps shown and fill and this is very
  • 00:21:34
    commonly a four mark question and an
  • 00:21:38
    exam so I would strongly recommend that
  • 00:21:40
    you learn best process now this is the
  • 00:21:43
    process for halogens but it's the same
  • 00:21:45
    mechanism as for a hydrogen halide the
  • 00:21:48
    only difference here is that as the
  • 00:21:51
    hydrogen so if halogen molecule is not
  • 00:21:54
    polar we need to get best polarity being
  • 00:21:57
    induced so as the the halogen molecule
  • 00:22:01
    gets close to the pi bond we get this
  • 00:22:04
    induced dipole being formed
  • 00:22:07
    meaning it's causing a delta positive in
  • 00:22:11
    a delta negative end and then the
  • 00:22:13
    mechanism is all the same so as i
  • 00:22:15
    this is a four marker regardless of
  • 00:22:17
    whether you're using a halogen or a
  • 00:22:19
    hydrogen halide so I would strongly
  • 00:22:21
    recommend that you do learn and memorize
  • 00:22:24
    these steps because they could be an
  • 00:22:26
    easy for Marx to pick up and then exam
  • 00:22:30
    now one thing that we do have to be a
  • 00:22:33
    weed of is that if we have an asymmetric
  • 00:22:35
    molecule we can get our B our molecule
  • 00:22:40
    from the HBR adding and two different
  • 00:22:43
    carbon atoms it could either add on to
  • 00:22:46
    carbon number one or carbon number two
  • 00:22:50
    so I'm going to make two possible
  • 00:22:53
    products now the two possible products
  • 00:22:56
    in this case are two bro more protein
  • 00:22:58
    and one bromo propane and one of them is
  • 00:23:01
    referred to as the major product and the
  • 00:23:04
    other is referred to as the minor
  • 00:23:06
    product and the reason that why we have
  • 00:23:09
    to label them as these as one is going
  • 00:23:12
    to be made more than the other and we
  • 00:23:14
    have to be able to explain that but it's
  • 00:23:16
    actually quite simple it is all to do
  • 00:23:20
    with the carbo cations that are formed
  • 00:23:22
    so if we have an asymmetrical alkene we
  • 00:23:25
    know that our carbon-carbon double bond
  • 00:23:27
    is going to have different atoms on
  • 00:23:29
    either side so we're forming these two
  • 00:23:32
    products and it actually means that we
  • 00:23:34
    form two possible carbo cations we can
  • 00:23:38
    either have the positive charge and the
  • 00:23:40
    carbon in the middle or we can have the
  • 00:23:42
    positive charge and a carbon at the end
  • 00:23:45
    if it is on a carbon that is an the
  • 00:23:48
    medal then we call this a secondary
  • 00:23:53
    carbo cation and if we have the positive
  • 00:24:02
    charge on the end then we call this a
  • 00:24:05
    primary carbo cation so if it's positive
  • 00:24:11
    and the center connected to to our cow
  • 00:24:13
    grips its secondary if the positive is
  • 00:24:15
    at the end only connected to one alkyl
  • 00:24:17
    group as a primary carbo cation
  • 00:24:21
    and when we have the secondary and
  • 00:24:26
    primary carbo cation they have different
  • 00:24:28
    level
  • 00:24:29
    of stability so the secondary carbo
  • 00:24:32
    cation has actually more stable than the
  • 00:24:34
    primary one because the charge can be
  • 00:24:37
    spread over the our car drips so if I
  • 00:24:40
    just go back we can see that that's
  • 00:24:42
    positive charge can be spread across the
  • 00:24:45
    to our characters whereas efforts the
  • 00:24:49
    primary can only spread and one
  • 00:24:52
    direction so the secondary has more
  • 00:24:56
    stable whereas the primary is less
  • 00:25:01
    stable and because of thus the major
  • 00:25:06
    product has formed from the more stable
  • 00:25:09
    carbo cation and the minor product is
  • 00:25:13
    formed from the less stable carbo cation
  • 00:25:21
    so to summarize how electrophilic
  • 00:25:24
    additions happen so we have them
  • 00:25:26
    proceeding via carbo cations these can
  • 00:25:29
    be primary second rate architects on it
  • 00:25:31
    I don't worry too much about the
  • 00:25:32
    tertiary just now um stability is going
  • 00:25:36
    to be the greatest for tertiary and the
  • 00:25:38
    least for priming it the carbo cations
  • 00:25:40
    are more stable when they have more
  • 00:25:42
    outer lips attached and the major
  • 00:25:44
    product is formed from the most stable
  • 00:25:47
    carbo cation so whether if you're
  • 00:25:51
    competing as a ter should and a second
  • 00:25:53
    ring then your ter shouldn't is going to
  • 00:25:55
    be the more stable if it's a second jay
  • 00:25:58
    at a primary then your primary is going
  • 00:26:00
    to be more stable and stable one will
  • 00:26:02
    always form your carbo cat will always
  • 00:26:05
    form your major product so let's finish
  • 00:26:10
    off by looking at some past paper
  • 00:26:12
    questions so for the first one we want
  • 00:26:15
    to look at how many second my bonds
  • 00:26:17
    there are and the organic compound with
  • 00:26:19
    the skeletal structure then some people
  • 00:26:21
    are quite good with skeletal structures
  • 00:26:23
    and they can just eat it right away if
  • 00:26:25
    not you may want to draw out so we've
  • 00:26:28
    got four carbons and between carbons 2 &
  • 00:26:34
    3 we have a double bond now we don't
  • 00:26:36
    show the hydrogen's on a skeletal
  • 00:26:38
    structure but we know that all the rest
  • 00:26:42
    are filled with hydrogens and for a
  • 00:26:47
    sigma bond is everywhere that we have a
  • 00:26:49
    single bond so we have 1 2 3 4 5 6 7 8 9
  • 00:26:55
    sorry 1 2 3 4 5 6 7 8 9 10
  • 00:27:00
    parties I have 10 single bonds but then
  • 00:27:04
    I also have a segment bond and my double
  • 00:27:06
    remember a double bond is made up of a
  • 00:27:09
    sigma and a PI so I have one Sigma bond
  • 00:27:14
    in my double giving me a total of 11
  • 00:27:17
    segment buds and then Bute 2 in reacts
  • 00:27:22
    with the certified potassium manganate
  • 00:27:24
    and what is the organic product so again
  • 00:27:27
    we're gonna draw it well we have beauty
  • 00:27:29
    in as the same molecule as above but I'm
  • 00:27:33
    just gonna draw a little cleaner down
  • 00:27:35
    here and always recommend drawing out
  • 00:27:40
    the structures if you're not sure and
  • 00:27:43
    what happens is I break my double bond
  • 00:27:45
    and I can only add on carbon number 2
  • 00:27:50
    and carbon number 3 so then my answer
  • 00:27:54
    has to be D there is no other possible
  • 00:27:57
    answer and then also from the January
  • 00:28:02
    2019 paper bromine reacts with propane
  • 00:28:05
    under normal lab conditions state type
  • 00:28:08
    and mechanism well this is going to be
  • 00:28:11
    an electrophilic addition reaction
  • 00:28:17
    because we have a halogen reacting with
  • 00:28:22
    an alkene and that is always
  • 00:28:24
    electrophilic addition and then draw the
  • 00:28:27
    structure well if we had propene we
  • 00:28:33
    would have our double bonds between
  • 00:28:34
    carbon 1 & 2 so therefore this is where
  • 00:28:39
    my bromine is going to add and then all
  • 00:28:45
    of my other bonds are all single bonds
  • 00:28:48
    to hydrogen and that is my reaction the
  • 00:28:54
    substance that is formed
  • 00:28:56
    and there you can see the mark skin
  • 00:28:58
    we're gonna look at one other question
  • 00:29:00
    which is from the June 2018 paper just
  • 00:29:03
    goes until a little bit more detail
  • 00:29:04
    about the mechanisms so boots are in has
  • 00:29:07
    two geometric isomers so we want to draw
  • 00:29:11
    them and name them so we know that we
  • 00:29:14
    can have in terms of skeletal we're
  • 00:29:16
    gonna have our double bond like that and
  • 00:29:20
    we can have both of the grips on the
  • 00:29:24
    same site or I can have one grip at the
  • 00:29:28
    top and one grip at the bottom and when
  • 00:29:32
    they're on the same side this is going
  • 00:29:34
    to be cess butte to in or you could
  • 00:29:38
    write Z good to in either one is
  • 00:29:43
    accepted and this one is going to be
  • 00:29:45
    trans be it to in or e Butte to even one
  • 00:29:54
    mark each for the structure and name of
  • 00:29:57
    your molecules and then part to explain
  • 00:30:01
    how geometric isomerism arises and butte
  • 00:30:03
    to in well we have restricted rotation
  • 00:30:13
    around the carbon-carbon double bond and
  • 00:30:17
    we have two different groups attached
  • 00:30:30
    to the carbon-carbon double bond so we
  • 00:30:33
    have a ch3 and a hydrogen and because
  • 00:30:36
    there are two different groups they can
  • 00:30:37
    destruction then they can be in
  • 00:30:39
    different positions for Part B for
  • 00:30:45
    marker as I mentioned earlier giving the
  • 00:30:47
    mechanism for the reaction between butte
  • 00:30:49
    2 in and hydrogen bromide using all of
  • 00:30:52
    the Carly arrows and all of the relevant
  • 00:30:54
    dipoles so the first thing we're gonna
  • 00:30:56
    do is we're gonna draw a beaut tui and
  • 00:31:01
    I'm gonna draw freely to split to make
  • 00:31:04
    it nice and easy you can draw it as
  • 00:31:07
    skeletal or you could draw as structural
  • 00:31:10
    that's completely up to you then we also
  • 00:31:13
    have our hydrogen and our bromide and we
  • 00:31:17
    know that this becomes Delta negative
  • 00:31:18
    and that becomes Delta positive I mean
  • 00:31:22
    that makes it clear so we have our
  • 00:31:25
    double bond Carly a dope going to the
  • 00:31:29
    hydrogen and then the bond between the
  • 00:31:32
    hydrogen and the bromine going to the
  • 00:31:35
    bromide and then that is going to form
  • 00:31:40
    our carbo cation intermediate so we have
  • 00:31:51
    bonded on the hydrogen and it leaves me
  • 00:31:53
    with this carbo cation and also leaves
  • 00:31:56
    me with the bromine which now has a lone
  • 00:31:59
    pair and is negatively charged and we
  • 00:32:02
    have this bromine so bromide attacking
  • 00:32:06
    and at this positive charge and then I
  • 00:32:10
    make my final structure that has got
  • 00:32:18
    that bromine atom added in like that so
  • 00:32:23
    you would get one mark for your first
  • 00:32:26
    Karla arrow from the double bond one
  • 00:32:28
    mark for second Carley arrow and the
  • 00:32:31
    polarity third mark is for your carbo
  • 00:32:34
    cation and then your fourth mark is for
  • 00:32:36
    showing the lone pair showing the arrow
  • 00:32:39
    and to the positive and also drawing
  • 00:32:41
    your product
  • 00:32:45
    to bromo butene so to bromo butane is
  • 00:32:49
    formed by the addition of the hydrogen
  • 00:32:51
    bromide to butte 1 and beautyrx in and
  • 00:32:53
    we want to explain why the atom economy
  • 00:32:56
    is going to be different for each
  • 00:32:58
    reaction so if we have the Butte to in
  • 00:33:03
    reaction that we've just seen there the
  • 00:33:06
    atom economy has 100% because we can
  • 00:33:11
    only make one products from that however
  • 00:33:16
    with Butte one in that says an
  • 00:33:23
    asymmetric molecule so not only do we
  • 00:33:28
    make some beautiful to bromo butan one
  • 00:33:32
    bromo butan as also for it and because
  • 00:33:39
    we get this other product we therefore
  • 00:33:42
    get a lower atom economy so we have to
  • 00:33:46
    take into account the fact that we have
  • 00:33:48
    a major and the manor product and you
  • 00:33:51
    have to specify endless particular
  • 00:33:53
    question that the other product is one
  • 00:33:55
    droplet in so there we can see our mark
  • 00:34:00
    schemes for the June 2018 paper now
  • 00:34:03
    that's everything for topic 5-8 alkanes
  • 00:34:06
    as I mentioned I'd strongly suggest you
  • 00:34:08
    practice drawing it the mechanism for
  • 00:34:10
    the electrophilic addition reactions
  • 00:34:12
    because it does come up as we've just
  • 00:34:14
    seen and if you have any questions
  • 00:34:16
    please feel free to leave a comment
  • 00:34:18
    below we hope to see you back soon
  • 00:34:20
    [Music]
Etiquetas
  • alcheni
  • isomerismu geometricu
  • reazioni chimiche
  • elettrofilica
  • ligami doppi
  • IES chimica
  • nomclature E/Z
  • nomenclature cis/trans
  • chimica organica
  • stabilità carbocatione