Alkynes

Question 1

alkynes

Answer 1

compounds containing a C=-C triple bond

	- sp hybridized
	- p orbitals make pie bonds
	- sp orbital makes sigma bond

Question 2

alkynes are similar to alkenes in their ability to function as

Answer 2

either a base or acid

Question 3

Nomenclature of Alkynes

Answer 3

same naming rules apply

- use yne at the end

Question 4

if a double and a triple bond are in equivalent positions then you list

Answer 4

double bond first but still name the final part with yne

Question 5

a lower pKa value is

Answer 5

higher in acidity

Question 6

acetylide ion

Answer 6

conjugate base of acetylene

- has a negative charge

Question 7

the equilibrium of an acid-base reaction favors the formation of

Answer 7

the weaker acid and weaker base

Question 8

greater stability leads to a weaker base and stronger acid

Answer 8

• an sp orbital will hold electrons closer to + leading to weaker base or stronger acid
  - sp3 will hold electron density farther away leading to a stronger base and weaker acid

Question 9

OH- cannot

Answer 9

deprotonate acetylene

Question 10

terminal alkynes are acidic and

Answer 10

can be deprotonated

Question 11

alkynide ion

Answer 11

the conjugate base of a terminal alkyne

Question 12

bases to deprotonate a terminal alkyne

Answer 12

• H2N-
  - H-
  - C4H9-
  - H-C=-C-

Question 13

bases that cannot deprotonate a terminal alkyne

Answer 13

• T-buO-
  - EtO-
  - OH-

Question 14

alkynes can be prepared from alkyl dihalides

Answer 14

• required two successive elimination(E2) reactions

- two E2 reactions because the alkyl halide will have two leaving groups

Question 15

germinal

Answer 15

a dihalide with both halogens connected to the same carbon atom

Question 16

vicinal

Answer 16

a dihalide with the two halogens being connected to adjacent carbon atoms

Question 17

both germinal and vicinal dihalides require two successive elimination reactions

Answer 17

• the second elimination reaction must be a very strong base

- sodium amide(NaNH2) dissolved in liquid ammonia(NH3)

Question 18

using NaNH2 dissolved in NH3 is frequently used for the preparation of terminal alkynes

Answer 18

the strongly basic conditions favor production of an alkynide ion which serves as the driving force for the overall process

Question 19

three equivalents of the amide ion are required to prepare an alkyne

Answer 19

• 2 for the success E2 reactions

- 1 to deprotonate the terminal alkyne and form the alkynide ion

Question 20

after the alkynide ion has formed and the reaction is complete, a proton source can

Answer 20

be introduced into the reaction vessel thereby protonating the alkynide ion to regenerate the terminal alkyne

Question 21

a terminal alkyne can be prepared by

Answer 21

treating a dihalide with excess(xs) sodium amide(NaNH2/NH3) followed by water(H2O)

Question 22

poisoned catalyst

Answer 22

a partially deactivated catalyst, a poisoned catalyst, it is possible to convert an alkyne into a cis alkene(without further reduction)

Question 23

lindlar’s catalyst

Answer 23

a common poisoned catalyst used to convert an alkyne into a cis alkene(without further reduction)

Question 24

using a poisoned catalyst does not produce

Answer 24

any trans alkene

Question 25

syn addition is used to form the

Answer 25

cis alkene as the major products

Question 26

explain the difference between NaNH2 and Na,NH3

Answer 26

• NH2- is a very strong base

- Na and NH3 are a source of electrons for reagents to use

Question 27

dissolving metal reduction

Answer 27

• alkynes can be reduced to trans by dissolving metal reduction
  - Na/NH3(l)

Question 28

radical anion

Answer 28

an intermediate with an unpaired electron instead of a pair of electrons

Question 29

to produce an alkane an alkyne can be treated with

Answer 29

H2 in the presence of a metal catalyst, such a Pt, Pd, or Ni

Question 30

to produce a cis alkene an alkyne can be treated with

Answer 30

H2 in the presence of a poisoned catalyst such as Lindlars catalyst of Ni2B

Question 31

to produce a trans alkene an alkyne can be treated with

Answer 31

sodium(Na) in liquid ammonia(NH3 l)

Question 32

Hydrohalogenation of Alkynes

Answer 32

• follows same rules as alkenes
  
  • markinokov addition
  • rate = k[alkyne][HX]^2
  • if H2O2 is used then the X will go to the least substituted position(like alkenes)