Orgo 1 Chp 4, 5, 6, 8 - Addition, Elimination, Substitution Rxns Flashcards Preview

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Flashcards in Orgo 1 Chp 4, 5, 6, 8 - Addition, Elimination, Substitution Rxns Deck (18):
0

Reagents: Alkane + heat (800 C)

Elimination
Dehydrogenation - Removal of 2 H's formation of an Alkene
Products: Alkene and H2
Mechanism not required

1

Reagents: "strong acid" (H2SO4) and heat

Elimination - E1
Dehydration of alcohols - Removal of OH and H to form an alkene
Products: Alkene + H2O
Regioselective: Zaitsev's rule applies - H is always taken from the B carbon with the fewest H's to form a more substituted alkene.
Stereoselective: Favors trans products

2

Regents: Alkyl Halide + Strong acid (EtOH)

Elimination - E1
Dehydrohalogenation - The removal of a halide and H to form an alkene
products: Alkene and Hydrohalide
involves formation of carbocation intermediate
Only occurs with secondary and tertiary halides
Follows zaitsevs rule - H removed from the most highly substituted carbon to create the most highly substituted alkene

3

Reagents: Alkyl Halide + "strong bulky base"

Elimination - E2
dehalogenation - removal of H and halide to form an alkene
Products: Alkene + Hydrohalide
capable with all degrees of halides
REQUIRES ANTI-COPLANAR ARRANGEMENT OF HALIDE AND BETA H. Overrides zaitsev's rule if not possible.

4

Reagents: Alkene + H2 + catalyst (Pt, Pd, Rh, Ni)

Addition
Hydrogenation - The addition of H2 across a double bond to form an alkane
Products: Alkane
H's are added by SYN ADDITION - to the less crowded side of the alkene.
Ex: addition to the BOTTOM of a bicyclic-methyl structure because of steric hinderance at the top. H's are more likely to be blocked by the large methyl groups on the top of the molecule and resort to addition on the bottom where there is no hinderance.

5

Reagents: Alkene + Hydrohalide (H-X) (HBr, HCl, HI)

Addition
Hydrohalogenation - addition of H-X over a double bond to create an alkylhalide
Products: Alkylhalide
involves formation of a carbocation intermediate
Follows Markovnikov's rule: H is attacked by the double bond and added to least highly substituted side, so carbocation forms on the more highly substituted carbon.
**Can involve rearrangements via methyl or hydride shifts**

6

Reagents: Alkene + 1. H2B6 or H3B-THF or BH3 (not typical) in solvent + 2. H2O2, OH-

Addition
Hydroboration-Oxidation: Anti-Markovnikov addition of H and OH across the double bond
Products: hydrated alkane (plus H and OH) in opposite to Markovnikov's rule.
*Do not need to know mechanism only reagents and products
**Occurs under SYN ADDITION
First step: double bond attacks the H and adds to most highly substituted side and BH2 is added to the least substituted side.
Second step: hydrogenperoxide and OH- substitutes the BH2 for OH -- hence OH is now added to the least substituted side.

7

Reagents: Alkene + dilute acid (50% H2SO4 and 50% H2O)

Addition
Hydration - the addition of H and an OH group across a double bond to form a hydrated alkane.
Products: Hydrated alkene with hydronium ion
*Obeys Markovnikov's Rule - hydronium ion is deprotonated by the double bond and is added to the least substituted side creating a carbocation on the more substituted side. Then hydronium ion attacks carbocation, creating an oxonium ion. H2O comes in and deprotonates the oxonium ion leaving an OH group creating another hydronium ion.
This is opposite to Beta elimination dehydration and can be reversed by concentration of acid. Strong acids undergo elimination of an alkane, whereas dilute acids undergo addition to an alkene.

8

Reagents: Alkene + X2 (Br2 or Cl2)

Addition
Halogenation - the addition of 2 halides across a double bond to produce an alkane with 2 added halides "vicinal dihalide"
Products: dihalide alkane
Can only be Cl2 or Br2 not F or I
Anti addition - double bond attacks one X, that X attacks the bond back-- halonium ion is formed, then other halogen attacks the more highly substituted carbon, in asymmetrical molecules, on the opposite side of the halonium ion.
*no carbocation intermediate. Only the formation of the positively charges halonium ion.

9

Reagents: Alkene + H2O + X2 (only Br or Cl)

Addition
Halohydration - the addition of a halogen and OH across a double bond to produce an alkane with OH and Halide "vicinal halohydrin"
Products: alkane with OH group and X group + H-X
*very similar to halogenation
*Anti addition of X and OH
Follows markovnikovs rule with the addition of OH to the more highly substituted side
first, double bond attacks the dihalide causing their bond to break and one halonium ion to form. then H2O attacks from the opposite side of the halonium ion at the more highly substituted carbon. The other X ion comes back and attacks the oxonium ion formed removing a H leaving a OH, causing the anti addition of OH to the more substituted carbon.

10

Reagents: Alkene + HBr + light (hv) or peroxide (H2O2)

Addition
Antimarkovnikov addition of HBr - adds H and Br over a double bond to create alkylhalide that is opposite to markovnikovs rule (The Br is added to the least substituted side and radical is formed on the more highly substituted carbon)
Product: Anti-Markovnikov alkylhalide
*Free radical mechanism - involving 3 steps
Initiation: the splitting of HBr into H-radical and Br-radical either by light or by the cleaving of H2O2 into 2 radicals.
Propogation: attack of double bond and Br radical to form a less substituted Br addition and a more highly substituted radical. Then H and Br cleave again adding a H onto the radical carbon, creating another Br radical.
Termination: The joining of two radical species to terminate the propogation of the rxn.

11

Reagents: Alkene + peroxy acid (RC=O OOH)

Addition
Epoxides - The addition of an epoxide ring to an alkene to form an alkane
Products: Epoxy alkane + carboxylic acid
*SYN ADDITION = CIS EPOXIDE
*No mechanism required, just reagents and products

12

Reagents: Alkene + O3 + Zn, H2O or (CH3)2S

Addition
Ozonolysis - the cleaving of alkene in half and the addition of O to end of each double bond
Products: the cleaving of alkene into two separate structures - can produce aldehydes or ketones
Aldehyde R-C=O -H or Ketone: R-C=O -R
*Do not need to know mechanism only reagents and products

13

Reagents: Secondary or Tertiary Alcohol Alkane + HX

Substitution SN1
halogenation of alcohols - OH group substituted by halide
Products: alkyl halide + H2O
*Contains a carbocation intermediate.
First step is the protonation of the OH by OH attacking the H and dissociating it from the X-- creating a oxonium ion and a halide ion
Second step is the dissociation of the oxonium ion to form a carbocation (slow step)
Third step is the attack of X- ion on the carbocation

14

Reagents: Primary alcohol alkane + H-X

Substitution Sn2
Halogenation of alcohols - The substitution of OH by halide
Products: primary alkyl halide + H2O
*undergoes substitution by BACKSIDE ATTACK -- THIS INVERTS STEREOCHEMISTRY AT THE a CARBON.
2 step mechanism
involves no carbocation intermediate
1. OH attacks the H kicking out the X- ion and creating an oxonium ion that is primary (can not leave by itself)
2. The remaining X- ion attacks the alpha carbon from the opposite side of the oxonium ion, attaching to the carbon and kicking off the OH (inversion of stereochem)

15

Reagents: Primary or Secondary alcohol alkane + SOCl2 + pyridine (or K2CO3)

Substitution
Thionyl Chloride - used on primary or secondary OH's to substitute for Cl
Products: alkylchloride + HCl + SO2

16

Reagents: any degree alcohol alkane + PBr3

Substitution
Phosphorous Tribromide - Substitution of an OH group with Br
Products: 3RBr + H3PO3

17

Reagents: Alkane + X2 (Cl or Br) + light (hv)

Substitution
Halogenation of Alkanes - the replacement of H's with either Cl (chlorination) or Br (bromination) halogens
Products: Alkane with Cl/Br replacement of H's
*Free-radical mechanism
*Highly reactive, will continue replacing H's until no more H's are present on the molecule
1. Bond between the dihalide cleaves producing two halide radicals
2. One halide radical attacks a H on the cleaving the H bond in half creating a H-X and a radical alkyl group
3. Radical alkyl group cleave another dihalide substituting one halide and creating another halide radical with the other
4. Termination occurs when two radicals combine.
*Br targets tertiary carbons only
*Cl is less selective, but is mainly only useful when all carbons have = amounts of H's or else Cl will attach to the most substituted carbon first.