alkena dan alkuna
DESCRIPTION
KIMIA ORGANIKTRANSCRIPT
ALKENA DAN ALKUNA
Alkane R–CH2–CH2–R CnH2n+2 This is the maximum H/C ratio for a given number of carbon atoms.
Alkene R–CH=CH–R CnH2n Each double bond reduces the number of hydrogen atoms by 2.
AlkyneAlkuna R–C≡C–R CnH2n-2 Each triple bond reduces the
number of hydrogen atoms by 4.
Alkena dan Alkuna
Alkena: Seny. hidrokarbon jenuh yg mengandung ikatan rangkap 2
Alkuna: Seny hidrokarbon tak jenuh yg mengandung ikatan rangkap 3
H2C CH2 H2C CHCH2CH3
Etena 1-ButenaSikloheksena Siklopentena
HC CH H3CC CCH3
Etuna 2-Butuna
IUPAC Rules for Alkene and Cycloalkene Nomenclature
1. The ene suffix (ending) indicates an alkene or cycloalkene. 2. The longest chain chosen for the root name must include both carbon atoms of the double bond. 3. The root chain must be numbered from the end nearest a double bond carbon atom. If the double bond is in the center of the chain, the nearest substituent rule is used to determine the end where numbering starts. 4. The smaller of the two numbers designating the carbon atoms of the double bond is used as the double bond locator. If more than one double bond is present the compound is named as a diene, triene or equivalent prefix indicating the number of double bonds, and each double bond is assigned a locator number. 5. In cycloalkenes the double bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule. 6. Substituent groups containing double bonds are: H2C=CH– Vinyl group H2C=CH–CH2– Allyl group
IUPAC Rules for Alkyne Nomenclature 1. The yne suffix (ending) indicates an alkyne or cycloalkyne. 2. The longest chain chosen for the root name must include both carbon atoms of the triple bond. 3. The root chain must be numbered from the end nearest a triple bond carbon atom. If the triple bond is in the center of the chain, the nearest substituent rule is used to determine the end where numbering starts. 4. The smaller of the two numbers designating the carbon atoms of the triple bond is used as the triple bond locator. 5. If several multiple bonds are present, each must be assigned a locator number. Double bonds precede triple bonds in the IUPAC name, but the chain is numbered from the end nearest a multiple bond, regardless of its nature. 6. Because the triple bond is linear, it can only be accomodated in rings larger than ten carbons. In simple cycloalkynes the triple bond carbons are assigned ring locations #1 and #2. Which of the two is #1 may be determined by the nearest substituent rule. 7. Substituent groups containing triple bonds are: HC≡C– Ethynyl group HC≡CH–CH2– Propargyl group
Tatanama Alkena dan Alkuna Mengacu sistem IUPAC, alkena diberi nama dg mengganti akiran –
ana dari alkana dg –ena (jika memp 2 ikatan rangkap maka disebut diena, sementara hidrokarbon dg 3 ikatan rangkap 2 disebut triena. Alkuna diberi akhiran –una
Rantai terpanjang: rantai utama yg mengandung ikatan rangkap 2 atau 3
Pemberian nomor dimulai dari C ujung yg paling dekat dg ikatan rangkap
H3CH2CHC CCH3
CH3
2-metil-2-pentenaH2C CCH CH2
CH3
2-metil-1,3-butadiena
H3CC CCH3
2-ButunaH3CC CCH2CH3
C CH
C C
H3C
H CH3
H
trans-2-butena
Sifat Fisika Alkena dan Alkuna
Alkena dan alkuna relatif lebih reaktif daripada alkana, namun tidak luar biasa polar
Dg demikian sifat fisiknya hampir sama dg alkana
Kepolaran alkuna > alkena > alkanaBaik alkana, alkena, dan alkuna tidak
larut dlm air
Sintesis Alkena Alkena dpt disintesis dari alkilhalida dan basa kuat melalui
reaksi eliminasi bimolekuler(E2)
HO HH2C CHCH3
Br
+Kalor
HOH + H2C CHCH3 + Br-
(CH3)2CHOHH2SO4
100 oCH3CHC CH2 + H2O
Mekanisme Reaksi Sintesis Alkena
CH
OH
H3C CH3
Tahap 1. protonasi
HCH
OH2
H3C CH3 -H2OHC
H3C CH3
Tahap 2. Lepasnya H+
HC
H2C CH3
HHC
H2C CH3
H
-H+
HC
H2C CH3
Sintesis Alkuna
RHC CHR
Br BrOR dalam etanol
-HBrRHC CR
Br
RHC CR
Br
NaNH2, KOH pekat + Kalor
-HBrRC CR
Alkuna dpt disintesis dari alkilhalida dan basa kuat melalui 2 reaksi eliminasi bimolekuler(E2)
Adisi Halogen Halida pada Alkena dan Alkuna
H2C CH2 + HX CH3CH2Xetil halida
HC CHHX
H2C CHXHX
CH3CHX2
H2C CH2H Cl
H2C CH2
H
+ Cl
H2C CH2
H
+ Cl H2C CH2
H
Cl
Reaksi adisi kebalikan dari reaksi eliminasi
Mekanisme reaksi adisinya (substitusi elektrofilik)
Refresh
H3CHC CH2HCl(adisi)
Aturan Markonikov
H3CHC CH2HCl
H3CHC CH2
Cl
H
bukan H3CHC CH2
H
Cl2-kloro propana
(H3C)2C CHCH3HBr
(H3C)2C CHCH3
Br
H
Dalam reaksi adisi HX kepada alkena tak-simetris, H+ dari HX menuju ke karbon berikatan rangkap yg telah lebih banyak memiliki hidrogen
Contoh
Penalaran Aturan Markonikov
H3CHC CH2
H+
H3CHC CH2
H
Keadaan transisi
H3CH2C CH2
Karbokation (C+) primer tdk stabil
H3CHC CH2
H+
H3CHC CH2
Keadaan transisi
H
H3C CH
CH3
C+ sekunder lebih stabil
Penalaran aturan Markonikov dilakukan dg mempelajari mekanisme reaksi adisi HX pada alkena, dg didasarkan pada kestabilan karbokation
Tingkat kestabilan karbokation tersier > sekunder > primer
H3CHC CH2
H+
H3C CH
CH3
ClH3CHC CH2
Cl
H2-kloro propana
(H3C)2C CHCH3 (H3C)2C CHCH3
Br
H
H+
(H3C)2C CH2CH3
Jadi
Adisi Halogen (Br2 dan Cl2) pada Alkena dan Alkuna
Jika Br2 dan ikatan pi mulai bergabung, elektron pi dapat menyebabkan Br2 terpolarisasi sehingga ujung molekul bromida sebagian positif dan ujung lainya (parsial) negatif
Langkah 1.
C C
R
R R
R
Br Br
C C
R
R R
R
Br
ion bromonium
+ Br
Langkah 2. masuknya Br
C C
R
R R
R
Br
Br
C C
R
R R
R
Br
Br
dibromo alkana
Adisi Halogen dan Air Jika alkena + Br2 ato Cl2 + air Camp. Adisi 1,2-
halohidrin contoh
H3CHC CH2 + Br2 + H2O H3CHC CH2
BrOH
1-Bromo-2-propanol
+ HBr
(1,2-bromohidrin)
Mekanismenya:
H3CHC CH2
Br Br
-BrH3CHC CH2
BrOH
H3CHC CH2
Br
OH
H
-H+
H3CHC CH2
Br
OH
Contoh soal Selesaikan reaksi berikut dg menunjukkan
intermediet bermuatan positif!
H3CHC CCH3
CH3
+ Br2CCl4a.
b. H3CHC CCH3
CH3
+ Br2 + H2O
Adisi H2SO4 dan H2O pada alkena dan alkuna H2SO4 berfungsi sebagai katalisator yg merupakan
seny asam pengadisi ikatan rangkap Mekanisme reaksinya:
H OSO3H H+ + OSO3H
H3CHC CH2 CH3CHCH3
OH
CH3CHCH3
OH
H
-H+CH3CHCH3
OH2-propanol
Contoh Soal Selesaikan reaksi berikut dan apa produk utamanya!
Tuliskan persamaan reaksi yg memperlihatkan bagaimana membuat senyawa berikut:
H3CH2CHC CH2 + H2OH+
a.
b. CH2CH3 + H2OH+
CH
OH
CH2CH2CH2CH3H3C