Electrophilic Addition to Alkenes and Alkynes Flashcards

1
Q

why is the C-H bond of an alkene shorter than in an alkane?

A

in the alkene, the electrons in the bond are held more closely to the carbon as it has more ‘s’ character.

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2
Q

what does it mean if something has more ‘s’ character?

A

orbitals with more ‘s’ character tend to be more spherical.

more tightly bound to the nucleus.

better at bonding with other atoms.

bond angle and strength are also impacted.

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3
Q

why is the boiling point of cis alkenes normally higher than trans alkenes?

A

carbon sp² - sp³ bond is polarised towards sp², causing a dipole moment.

in cis, these reinforce, in trans they cancel out.

larger dipole moments = larger dipole interactions = larger boiling points.

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4
Q

what determines the acidity of alkenes?

A

stability of resulting carbocation after double bond has been protonated.

more stable carbocation = more acidic alkene.

protonation forms + charge on one carbon in the db, = carbocation intermediate.

stabilised by resonance, positive charge is delocalised over adjacent π bonds.

alkenes w/ electron-withdrawing groups stabilise + charge and increase acidity.

alkenes w/ electron-donating groups destabilise the + charge and decrease the acidity.

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5
Q

why are alkenes more acidic than alkanes?

A

when H is removed, a carbanion is formed, stabilised by the π bonds.

The bond is able to delocalise the - charge, distributing it over a larger area.

in contrast, the alkane conjugate base is destabilised by inability of sp³ to accomodate additional - charge.

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6
Q

what are the key differences between the IRS of alkenes and alkanes?

A

C-H stretch in alkene is >3000, but lower in alkanes.

C=C stretch in alkenes is between 1620-1680.

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7
Q

what are the differences between the IRS of alkenes and alkynes?

A

C-H stretch in alkenes = 3000-3100,
but larger in alkynes.

C=C stretch in alkynes is >1900.

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8
Q

tell me about the bromination of alkenes?

A

Br₂ + alkene -> Br
\__
\
Br

toluene is the solvent.

alkenes turn colourless.
alkanes stay orange/brown.

alkene = Nu
(HOMO is π filled).

Br = E
(LUMO is σ* filled).

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9
Q

why do more electron rich alkenes lead to faster rates of bromination?

A

alkyl substituents raise the energy of HOMO C=C bond by hyperconjugation.

electronic effects override the steric effects.

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10
Q

what is Markovnikov’s rule?

A

the addition of a protic acid (HX) to an alkene, the acid H becomes attached to the C w/ the fewest substituents.

The X attaches to the C w/ more alkyl substituents.

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11
Q

what are some examples of naturally occurring and synthetic alkenes?

A

natural = retinal and cholesterol.

synthetic = polyethene and terpenes.

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12
Q

what part does regioselectivity play in Markovnikov’s rules regarding addition reactions between alkenes and HBr?

A

HBr + alkene proceeds w/ carbocation intermediate, + charge is sp² hybridised.

empty p-orbital perpendicular to plane of molecule.

stability of carbocation is reinforced through hyperconjugation and inductive effects.

favours more substituted alkyl bromide as major product.

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13
Q

alkenes react with peracids to give what?

A

epoxides.

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14
Q

epoxides can be opened with water under acidic conditions to give what?

A

trans diols.

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15
Q

why do more substituted alkenes epoxidise faster?

A

higher electron density around db.

π electrons of double bond held more tightly, leading to stronger interaction with peroxide agent.

steric hindrance around db in more substituted alkenes is reduced.

means peroxide reagent can approach the db more easily.

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16
Q

what is anti-markovnikov addition?

A

when the H binds to the most substituted carbon atom and the incoming group binds to the least substituted group.

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17
Q

why is a tertiary cation more stable than a primary one?

A

3⁰ has charge spread out over a larger area.

alkyl groups provide steric hindrance around C atom.

also has resonance stabilisation with π bonds.

18
Q

how does one make the antimarkovnikov product?

A

hydroboration-oxidation.

alkene + BH₃ = boron intermediate, can react with water = alcohol product.

boron intermediate can be attacked by water with oxygen peroxide presence to form alcohol.

19
Q

what is special about peroxides?

A

they can stimulate homolytic fission.

peroxides have a weak O-O bond that can be broken by heat or light, making 2 radicals.

radicals can then initiate and propagate.

20
Q

tell me about the structure and stability of radicals:

A

resonance stability

inductive effects

hyperconjugation

21
Q

why is a tertiary radical more stable than a primary radical?

A

alkyl groups attached to the carbon provide more electron density and inductive stability.

branching results in steric hindrance.

22
Q

why are allyls more stable than phenyls and vinyls

A

π electrons in the allyl system can delocalise over the entire conjugated system, provides stability.

π electrons in the phenyl group are delocalised over the ring system, less stable.

23
Q

What type of fission does peroxide undergo?

A

Homolytic

24
Q

Why is a primary radical not feasible in propagation?

A

There is a greater electron inducting effect in tertiary.

Tertiary have more alkyl groups, stabilises unpaired electron pairs and reduces reactivity

25
Q

What is the product of high concentration of Br2 for radical bromination?

A

Dibromide

26
Q

What is the product for low concentration of Br2 for radical bromination?

A

Allylic or benzylic bromination

27
Q

What happens when the allylic radical is formed?

A

It will react with bromine to form the desired product. This reaction is thermodynamically favoured.

28
Q

How is the allylic radical stabilised?

A

Can undergo resonance, delocalised electron into pi system.

Hyperconjugation through interaction between sigma orbital and pi orbital

29
Q

Why are allylic C-H bonds weaker than normal C-H bonds?

A

Due to resonance of stabilisation of radical

30
Q

what are the the conditions for the reduction of alkenes?

A

alkene hydrogenation - metal catalyst on solid support.

carbonyl reduction - HaBH⁴ + H₃O(+).

31
Q

what are the conditions for a less hindered alkene?

A

Pd catalyst at RT, 1 atm.

further reaction is 70⁰C and 200 atm.

32
Q

what are the conditions for the reduction of alkynes?

A

Pd/C.

Lindlar catalyst.

CaCO₃.

Li, NH₃.

33
Q

what determines the reactivity towards H in hydrogenolysis?

A

bond strength.

electron-withdrawing groups can destabilise the bond being cleaved.

steric hindrance.

catalyst used.

34
Q

what do substituents on the aromatic ring influence?

A

rate and regioselectivity of electrophilic aromatic substitution reactions.

35
Q

why do inductive electron withdrawing groups make compounds less reactive?

A

exert withdrawing effect through sigma bonds.

pulls electron density away from atoms.

electron density at reactive site is less extreme, less available for bonding or reacting.

aromatic ring is electron poor.

36
Q

what are the product ratios of ortho meta and para for the substitution of aromatic rings?

A

Ortho and para products are favored due to resonance stabilization. The electron-donating groups (EDGs) enhance the ortho and para directing effects, while electron-withdrawing groups (EWGs) can direct substitution towards the meta position.

37
Q

why does N,N-dimethylanaline have a faster rate than methoxybenzene and benzene?

A

N,N-dimethylaniline contains a dimethylamino group (-N(CH3)2) as an EDG, which can donate electron density to the aromatic ring through resonance effects. This electron-donating nature of the dimethylamino group increases the electron density on the aromatic ring, making it more nucleophilic and facilitating reactions.

38
Q

activating vs deactivating:

A

EDG - para, ortho (activating).

EWG - meta (deactivating).

39
Q

how does deactivating work?

A

EWGs direct substituents towards meta.

withdraws electron density from ortho and para, makes + charge less stable.

E- is more likely to attack meta, where electron density is higher and + charge is stabilised.

40
Q

what is unusual about halobenzenes?

A

they react as both EWG and EDG.

41
Q

why does meta bromobenzene not have stability?

A

no stabilisation due to donation of lone pair on bromine.

42
Q

what is the difference between inductive effects and conjugation?

A

Inductive effects occur through σ bonds, electronegative atom pulls electron density towards itself.
partial - charge made.

conjugation is π electrons in p-orbitals, delocalised over a conjugate system.
overlap of p-orbitals above and below plane of molecule.
leads to resonance stabilisation and distribution of electron density over a larger portion of the molecule.