ALKYNES ANSWERS
1
a)2,5-dimethyl-3-hexyne
b)3,3-dimethyl-1-butyne
c)2,4-octadien-6-yne
d)3,3-dimethyl-4-octyne
e)2,5,5-trimethyl-3-heptyne
f)6-isopropylcyclodecyne
g)2,5-octadiyne
h)3,6-dimethyl-2-hepten-4-yne
i)1,3-hexadien-5-yne
j)3,3-dimethyl-1,5-hexadiyne
2
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ALKYNES ANSWERS
ALKYNES ANSWERS
a)
b)
c)
d)
e)
f)
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ALKYNES ANSWERS
ALKYNES ANSWERS
3yikes!
a)(3E,5E,11E),-1,3,5,11-tridecatetraen-7,9-diyne
b)1-tridecen-3,5,7,9,11-pentayne
4A terminal alkyne has a triple bond on the end carbon in a chain. The alkyne triple bond is not on the last carbon in a chain in an internal alkyne.
5
a)
b)
c)
6
a)
b)
c)
7
8
a)
b)
c)
9
a)
b)
10
a)
b)
11 A reaction is said to go to completion when Keq 103 (pKeq -3). For a reaction to proceed at least 80% we would expect pKeq to be not more than –1. Use this value in the equation for pKeq and solve for the strength of the base required, i.e., solve for pKb.
(pKeq = pKa + pKb –14) (-1 = 19.3 + pKb - 14) (pKb = 14 – 1 – 19.3) pKb = -6.3
From this calculation, the pKb of the base must be -6.3 to deprotonate acetone at least 80%.
The pKb values of the bases listed can be calculated from the pKa values given for their conjugate acids using the formula (pKa + pKb = 14) which is true for all conjugate acid/base pairs. The pKb values of the bases are:
pKb of KOH = -1.74, pKb of Na+-:CCH = -11, pKb of NaHCO3 = 7.6, and pKb of NaOCH3 = -1.6
Of these four bases, only b) [sodium acetylide] is strong enough (pKb = -11) to remove an H+ ion from acetone, since it is a very weak acid.
12
a)
b)
c)
d)
13
a)
b)
c)
14
15
16
a)C8H10 vs. C8H18 8H/2 = 4DU
b)1 triple bond (since only 1 equiv. H2 is absorbed over Lindlar catalyst)
c)1 double bond (since 3 equiv. H2 are absorbed over Pd catalyst and one of these equiv. is for one bond in the triple bond, leaving a second bond in the triple bond and one bond not in the triple bond, i.e., must be from a double bond)
d)1 ring (since hydrogenation only accounted for 3DU but the formula indicates there is one more DU.
e)
17
a)
b)
c)
d)
e)
f)
18
a)
b)
c)
d)
e)
19
a)
b)
20
a)
b)
c)
d)
e)
21Both alkenes and alkynes are hydrated (hydration = addition of H2O) to a Markovnikov product without rearrangement using Hg+2 catalyst, H2O, and H2SO4.
In both cases, the reaction is initiated when Hg+2 ion (a strong electrophile) adds to the less substituted carbon in the double or triple bond.
In both cases, H2O adds next to the carbocation (that formed on the more substituted carbon).
In both cases a bridged mercurinium cation intermediate forms preventing rearrangement and ensuring antiaddition of Hg+2 and H2O.
With alkenes, hydration saturates the double bond and NaBH4 is required in a final step to replace the Hg+2 ion with H. With alkynes, hydration produces an unsaturated product, an enol, which tautomerizes to a ketone. H+ ion from H2SO4 replaces Hg ion so the addition of NaBH4 is not required.
22 Both alkenes and alkynes give positive results for the lab tests for unsaturation, i.e.,
Orange Br2 is decolorized as it adds to alkenes and alkynes
Purple KMnO4 is decolorized as it oxidizes alkenes and alkynes (in neutral or alkaline media, brown MnO2 precipitates)
Both alkenes and alkynes dissolve exothermically in conc. H2SO4 as the acid adds to the bond forming a bisulfate.
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ALKYNES ANSWERS