Alkenes, Alkynes, and Elimination Rxns

Question 1

Higher melting point, cis or trans alkenes?

Answer 1

Trans

Question 2

Higher boiling point, cis or trans alkenes?

Answer 2

Cis

Question 3

What are the two types of elimination reactions (not E1 and E2)

Answer 3

Dehydrohalogenation and dehydration

Question 4

What are the conditions required for dehydration?

Answer 4

Acid and high heat

Question 5

What conditions favor E1?

Answer 5

Same as SN1: polar protic solvent, stability of the carbocation, highly substituted carbon, good leaving group, absence of good nucleophile.

Question 6

How to favor E1 over SN1?

Answer 6

Higher temperatures favor E1

Question 7

What types of products are favored in E2?

Answer 7

Bond will form on more highly substituted carbon and trans isomer will predominate.

Question 8

How to favor E2 over SN2?

Answer 8

Steric hindrance is key, so highly substituted carbons and bulky bases make it harder for the SN2 to work. Also, a stronger base, because it will tend to tear off B-hydrogen before it reaches the a-carbon.

Question 9

Alkene + (H2 + Pd/Pt/Ni) –> ?

Answer 9

Syn addition of H (reduction) to the double bond.

Question 10

CH3CH=CH2 + HX –> ?

Answer 10

CH3-CXH-CH3 (Proceeds through most stable carbocation intermediate when the H+ adds to the less substituted carbon in the bond)

Question 11

CH3CH=CHCH3 + Br2 –> ?

Answer 11

CH3CBrH-CBrHCH3 (Anti-addition proceeds through cyclic halonium ion intermediate. First double bond grabs one Br, leaving another Br-, forms a cyclic halonium, then Br- attacks attacks the ion on the opposite face.)

Question 12

CH3-CH2OH + Acid + High heat –> ?

Answer 12

H2C=CH2 + H2O (Through E1 and E2 mechanisms. E2 if the OH is terminal since that would not form a stable cation.)

Question 13

H2CCH=CH2 + Acid + H2O + low heat –> ?

Answer 13

H3C-CHOH-CH3 (Through carbocation intermediate).

Question 14

H3CCH=CH2 + uv, peroxides, or oxygen + HBr –> ?

Answer 14

H3CCH2CH2Br + Br radical (Br free radical steals double bond and makes the secondary carbon a free radical [most stable], then the free radical attacks HBr and steals the H, leaving a Br free radical).

Question 15

Alkene + BH3 + THF….. + H2O2 + OH- –> ?

Answer 15

Syn alcohol alkane

Question 16

Alkene + cold, dilute KMnO4 –> ?

Answer 16

Syn vicinal diol (glycol) + MnO2(s)

Question 17

Alkene + 1) KMnO4, OH-, heat and 2) H+ (acid wash) –> ?

Answer 17

Alkene cleaved at double bond and carbons oxidize as much as possible (single carbons become CO2, primary carbons become carboxylic acid, secondary carbons become ketones)

Question 18

Alkene + 1) O3, CH2Cl2 2) Zn/H2O or (CH3)2S –> ?

Answer 18

Alkene cleaved at double bond and carbons oxidize to ketone or aldehyde.

Question 19

Ozonolysis + mild reducing agent (NaBH4 or LiAlH4) –> ?

Answer 19

Aldehydes and ketones from ozonolysis reduced to alcohols

Question 20

Alkene + CH3CO3H (peroxyacetic acid) or mCPBA (m-chloroperoxybenzoic acid) [these are both peroxycarboxylic acids] –> ?

Answer 20

Epoxide created between the carbons of the double bond.

Question 21

Alkene + radical, heat, high pressure –> ?

Answer 21

Polymerization

Question 22

Common name for ethyne

Answer 22

Acetylene

Question 23

What’s a special property of terminal akynes?

Answer 23

The acidity of the terminal H, with pKa of 25

Question 24

Two ways to synthesize alkynes?

Answer 24

1) Two rounds of elimination with geminal or vicinal dihalide with high heat and strong base.
2) Adding triple bond to existing haloalkane using strong base to pull of acidic hydrogen, which makes nucleophilic attack on the haloalkane.

Question 25

Two ways to reduce alkynes to alkenes?

Answer 25

1) Lindlar's catalyst: palladium on BaSO4 with quinoline. Produces cis isomer. 2) Na in liquid NH3 (<-33 C). Produces trans isomer and proceeds through free radical mechanism.

Question 26

Propyne + Br2 --> ?

Answer 26

Trans dihaloalkene via cyclic halonium ion intermediate.

Question 27

1-Butyne + HX + uv/peroxide --> ?

Answer 27

mostly trans 1-haloalkene

Question 28

Explain hydroboration of alkyne

Answer 28

3 Alkynes + 1/2 B2H6... boron coordinates with three aklynes and followed by acetic acid wash makes 3 cis aklenes. If terminal alkynes, disubstituted borane used to prevent further boration of vinyl intermediate to alkane. Instead, vinylic borane intermediate oxidatively cleaved with H2O2 to make vinyl alcohol with tautomerizes to more stable aldehyde.

Question 29

Alkyne + hot basic KMnO4 --> ?

Answer 29

Just like with alkenes, triple bonds are cleaved and carbons maximally oxidized.

Question 30

Alkyne + ozone + H2O --> ?

Answer 30

Triple bond cleaved and carbons maximally oxidized

Question 31

Explain addition of HX to an alkene

Answer 31

Electrons of double bond act as Lewis base and attack the H of HX creating a carbocation. The X- then attacks the carbocation to form a haloalkane.

Question 32

Explain addition of X2 to alkene

Answer 32

A rapid process where the double bond acts as nucleophile and attacks X2, where X acts as the leaving group, forming a cyclic halonium ion intermediate. X- then attacks the opposite side for an anti-addition, forming trans dihaloalkane.

Question 33

Explain addition of H2O to alkene

Answer 33

Performed under acidic conditions at low temperatures because at high temperatures the opposite reaction is favored. Double bond acts as nucleophile and attacks free H+ in solution, forming carbocation. Water attacks the carbocation and drops its H+ back into solution.

Question 34

Explain the E1 reaction mechanism

Answer 34

A good leaving group leaves and forms a stable carbocation. A base attacks the hydrogen on neighboring carbon and the two electrons left from that bond move to form a double bond with the carbocation. This competes with the SN1 reaction, but is favored over it with high heat.

Question 35

Explain the E2 reaction mechanism?

Answer 35

A strong, bulky base attacks the hydrogen of a carbon adjacent to a highly substituted carbon with a good leaving group under high temperatures. In one swoop, the base extracts the hydrogen, whose electrons move to make the double bond, and the leaving group leaves with the electrons from its bond. The trans isomer and the more heavily substituted carbon double bond will be favored.