Revision:Further Organic Chemistry - H (Higher)

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H.1 Stereoisomerism

H.1.1

Geometric isomers occur because the C=C bond is not able to rotate. Thus, cis and trans isomers are possible when there are different compounds possible if the 'top' and 'bottom' ranches are swapped on either side. For things like 1,2-dichloroethane, the cis isomer is when the two chlorines are both top (or bottom) where as the trans is where they are opposite...because cis means on the same side, and trans means on opposite sides.

H.1.2

It is possible for one of the cis or trans isomers to be polar while the other is not, and this allows them to have different boiling points. In 1,2-dichloroethane, the cis-isomer is polar, and so it's boiling point is 60 as opposed to 47 for the trans isomer.

Chemical properties - When cis-but-2-ene-1,4-dioic acid is heated, H₂O is eliminated and the two carbons at either end of the chain are linked across the remaining central oxygen, giving 2-butene-1,4-dioic anhydride (though I don't think we need to know the name). The trans isomer will not react, because the alkanoic acid bits are on opposite sides rather than together.

H.1.3

Rotation can also be stopped by the carbon atoms in question being linked together as part of a loop (a cycloalkane). For example, 1,2-dichloro cyclobutane has both cis and trans isomers because the ring stops the bond between the carbons with the Cl molecules attached from rotating. The same goes for cyclopropane.

H.1.4

Plane polarised light is light which only travels in one plane (see physics waves & optics sections for more info). Enantomers (or optical isomers) can be determined by their effects on such light. There are, for an optically active compound, two different enantomers, which will rotate plane polarised light passing through them in opposite directions. The direction of rotation can be determined using an analyzer, or a polaroid.

H.1.5

Enantomers occur due to the fact that carbon atoms form four bonds. If the carbon is bonded to 4 different groups, then there will be two possible arrangements of them which are non-superimposable...you can demonstrate this with models, but just trust me. The two enantomers are generally denoted L and D-enantomers, but I don't think we need to know which is which. These are known as optical isomers because one is the mirror image of the other.

H.1.6

A racemic mixture is a substance which contains equal percentages of both enantomers, and so has no effect on plane polarise light. Enantomers usually exist as racemic mixtures, though in many biological systems, one is favored over the other.

H.1.7

Physical properties are effectively identical for both enantomers, as are the chemical properties, except in the presence of other optically active compounds. This is particularly common in biological systems...for example, the tastes sweet and sour are optical isomers of the same compound.

H.2 Free radical substitution reactions

H.2.1

Free radicals are molecules with one (or more) unpaired electrons, and are very reactive...I'll be representing them as Cl° (with a bullet point)...

Alkanes with halogens - In the presence of UV light:

,

because the light provides sufficient energy to split the bonds (this is initiation). The Cl° then reacts with CH₄ as follows:

(By the way, everything here is in the gas state, so I won't write it each time)

The CH₃° then reacts with Cl₂, as follows:

this, and the previous are the propagation steps. This can be repeated to form CH₂Cl₂, CHCl₃ and eventually CCl₄.

Termination occurs when two free radicals react:

,

,

.

Methylbenzene with halogens - The halogen substitutes preferentially into the side chain (the methyl bit), with the same mechanism as above.

H.2.2

The effect of RCl on O₃ - As above, in UV light, the RCl bond breaks homolytically, forming CH₃° and Cl°. (this is initiation...)

followed by:

This produces more chlorine, thus producing a chain reaction.

Termination:

,

.

H.3 Electrophilic addition reactions

H.3.1

The reaction mechanism for electrophilic addition really needs a picture, but I'll try to describe it ... this example is hydrogen chloride adding to ethene, but could apply to anything. The reaction can be drawn with curly arrows...the Pi electrons from the ethene double bond move on to the less electronegative element in the electrophile (in this case hydrogen). This causes the electrons from the H-Cl bond to be moved onto the Cl atom. This creates CH₃C + H₂ and Cl^-, followed by which, the Cl^- is attracted to the carbocation and forms a dative covalent bond with the carbon. The movement of electrons should be drawn with curly arrows

H.3.2

Markolnikov's rule is basically that the less electronegative atom in a molecule adds to the atom carbon atom on the double bond with the most hydrogens already on it. This can be explained by the stability of the carbocation intermediate...either way, after the first atom adds on, there will be a carbocation, but if there are more electron donating (ie not hydrogen) groups around it, it will be more stable...it is the less electronegative atom which bonds first because it is attracted to the electron rich Pi electron area, and because the carbon will be stronger electron donating. (of course, this only applies to the major product, some of the other will still be produced)

H.3.3

Carbocations are more stable the more electron donating group they have around them, so the more CH₃ groups replace the carbons around a carbocation, the more stable it will be. Ph+ (which is a carbocation in a benzene ring) is more unstable than any alkane carbo-cations due to the loss of the delocalized Pi electron system.

H.4 Electrophilic substitution reaction

H.4.1 : The nitration of benzene

ie adding NO₂ to C₆H₆.

First a nitrating mixture is used to produce NO₂⁺. H₂SO₄ is added to HNO₃ and since H₂SO₄ is the stronger acid, and so it donates a proton to the HNO₃, producing H₂O⁺-NO₂.

This compound then decomposes to form H₂O + O=N⁺=O. This NO₂⁺ is then attracted to the delocalized Pi electron system of the benzene ring. The benzene ring donates an electron pair to form a dative covalent bond with the NO₂, but since there is already a hydrogen connected to the carbon, a carbocation is produced. Since this is unstable, the hydrogen will break off (and conveniently go back to reform the H₂SO₄, making it a catalyst) leaving Nitrobenzene.

H.4.2 : The production of methylbenzene with a Friedel-Crafts catalyst

The catalysts are FeCl₃ or AlCl₃, and these are used to polarise the electrophile. The reaction begins with a Haloalkane...

.

First the R-Cl bond is broken heterolytically with both electrons going to the chlorine, which then joins to the catalyst as above.

The R⁺ is then attracted to the delocalized Pi system, forming an intermediate where both the R and H are bonded to a carbocation, and then the H⁺ breaks off, and reacts with the Chlorine we left on the catalyst to form HCl., and we're left with alkylbenzene. It should be noted that the two steps happen sort of simultaneously, as the chlorine is pulled off by the catalyst the R forms a dative covalent bond with the benzene, so the mechanism should be drawn in one step.

H.4.3

Directing and rate effects...General principles

Electron donating groups ... Increase in delocalized electron density -- faster electrophillic substitution. These groups are also generally 2,4,6 directing...these groups include CH3, OH-.

Electron withdrawing groups ... decrease electron density -> slower electrophillic substitutions, and tend to be 3,5 directing...this table explains what we need to know...

As you can see, -Cl is deactivating, but 2,4,6 directing...that's why they're only general principles ;)

Oh...and the directing thing...that means that subsequent electrophillic substitutions will go in these positions if the one in question is on the 1 carbon.

For example:

(with the chlorines in the 2,4 and 6 positions)

H.5 Nucleophillic addition reactions

H.5.1

HCN first breaks to form H⁺ + CN^-. The CN^- molecule then attacks the central carbon atom in an alkanone (the one with the double bond to the oxygen...alkanals can also be used). This then forms N C-C(R)(R/H)-O- + H⁺ ... the H⁺ being the one broke off at the start, which then forms N -C-C(R)(R/H)-OH. This is then reacted as follows:

Which is forming carboxillic acid.

The rates may be relevant...Alkanals are faster than alkanones because there are fewer electron donating groups in alkanals, resulting in a larger charge, and so the CN^- nucleophile is more attracted. If one of the R groups is benzene, this is even slower as the becomes delocalized.

H.6 Nucleophillic Substitution reactions

H.6.1

The rate of nucleophillic substitution is affected by the polarity of the nucleophile and the polarity of the charge on the atom which is being attacked. For example, substitutions with OH^- will be faster than those with H²O, because of the greater polarity of the oxygen atom in OH^-.

The rate of substitution of CH₃NH₂ will be faster than NH3 due to the electron donating effect of the CH₃ group creating a larger charge on the N atom (which is the nucleophile).

H.6.2 : Inductive and steric effects.

Inductive: If an electron withdrawing group (such as a halogen) is connected to the carbon being attacked, then the charge will be increased, resulting in a greater rate of nucleophillic substitution. Conversely, if an electron donating group (such as CH₃) is attached, then the charge will be decreased, resulting in a slower rate.

Steric effects: This refers to the incoming nucleophiles being blocked by other groups surrounding the carbon atom being attacked. For example, R₃-C-X < R₂-CH-X < R-CH₂ < CH₃-X

Beyond the inductive effects, the bulky alkyl groups around the the first block the attack, making it difficult for the nucleophile to get to the C atom...and so on up the list.

Nb: SN₁ and SN₂ mechanisms apply here, and substitution rate into CH₃F < CH₃Cl < CH₃Br < CH₃I -it's like this because the C-I bond has the lowest bond energy, and so on up, and this outweighs the increased electronegativity (-> more delta +ve C with fluorine etc).

H.6.3

Benzene-X (a halogenated benzene ring) is unreactive to nucleophillic substitution reactions under normal conditions. The lone pairs on the halogen atom interact with the delocalized Pi system, which strengthens the C-X covalent bond. The high electronegativity of the ring also tends to repel any approaching nucleophiles ... The reactions are possible, however, at high temp and pressure (possibly with a catalyst)...ie

.

H.7 Elimination reactions

H.7.1 : Elimination of water from alkanols (dehydration)

The object here, is to produce an alkene from an alkanol, eliminating water in the process. This can be done in acidic conditions and high temperature, as follows.

The sulphuric acid propanates the alcohol.

Water is eliminated, leaving a carbocation...

The proton is reproduced, which allows it to rejoin the O--SO₂-O-H, making the acid a catalyst.

The more alkyl groups which surround the carbon atom, the faster this will occur, because the carbocation will be more stable ... therefore, tertiary alkanols are faster than secondary which are faster than primary.

H.7.2

The elimination of HBr from a bromoalkane can occur via two different mechanisms, one of which involves a nucleophillic attack.

E.1

The C-Br bond breaks spontaneously, and hetrolytically (with Br getting both electrons as follows:

this is then followed by the alkane forming an alkene and losing a hydrogen in the process:

this will occur most quickly with tertiary halogenoalkanes, because the carbocation will be more stable.

E.2

An OH^- nucleophile forms a dative covalent bond with one of the hydrogen atoms. This causes the hydrogen to be removed, leaving the carbon with a spare par of electrons, which then form a double bond with an adjacent carbon (which conveniently has a Br atom on it). This double bond formation causes the C-Br bond to break hetrolytically with the Br getting both electrons, thus (this all occurs in one step, but with curly arrows)

which is again eliminating HBr.

H.8 Addition-elimination reactions

H.8.1

Reaction of 2,4-dinitro phenylhydrazine ( H₂N-NH-Benzene-(NO₂2)(NO₂) ) with alkanals and alkanones.

Then water is eliminated:

This product is a yellow/orange precipitate with a characteristic melting point, often used to determine the alkanol/alkanone we started with.

H.9 Acid-base reactions

H.9.1

Phenol (a benzene ring with an OH on it) is acidic due to the stability of the Benzene-O^- ion (formed when it donates a proton). (This is all opposed to alkanols which are not significantly acidic). The acidity of phenols is due to the fact that the spare electrons around the O^- ion become involved in the delocalized Pi electron system, which makes the charge delocalized rather than concentrated on the O atom, which is energetically favorable, and more stable ... Phenols are still not strong acids though (weaker than RCOOH) - they do not react with Na₂CO_3(aq), but do react with NaCl - their Ka is about 10^-10.

If phenol has electron withdrawing groups substituted (ie NO₂), however, this again decreases the negative charge on the O^-, thus making them stronger acids...2,4,6 trinitrophenol has a Ka of about 10^-4. Electron donating groups, however, will make for weaker acids.

H.9.2 : Substituted alkanoic acids

Again, electron withdrawing groups -> increased bond polarity -> reduced charge on the resulting ion -> more stable, thus a stronger acid ... electron donating -> decreased bond polarity -> weaker acid.

H.9.3 : Ammonia, amines and amides

The basic nature of all these compounds comes from the fact that nitrogen has 5 valence electrons. 3 of these form covalent bonds, and two are left in a lone pair, which is capable of forming a dative covalent bond with protons.

Ammonia and amines

Amines are really just ammonia with R groups replacing the H atoms, and so they fit into a series. Each time another R group is added, the electron density donating effect creates a larger <altex>\delta ^-</latex> charge on the N atom, making it better at attracting and bonding with protons ... This means that Ammonia is weaker than primary amides, which are weaker than secondary amides ... unfortunately, tertiary amides fall between primary and secondary, due to the the fact that the increasing non-polar alkyl groups make tertiary amides less soluble in water, which more than accounts for the increased inductive effect - which is a pain, but you'll just have to remember it. An phenol amide is weaker than all of them because the lone pair becomes part of the delocalized Pi system.

Amides - These are very weak bases due to the electron withdrawing effect of the C=O group, which reduces the negative charge on the N atom, making it less able to attract protons, and less stable once it has one.

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