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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 tidak jenuh yg mengandung ikatan rangkap 2

� Alkuna: Seny hidrokarbon tak jenuh yg mengandung ikatan rangkap 3

H2C CH2 H2C CHCH2CH3

Etena 1-Butena

Sikloheksena Siklopentena

HC CH H3CC CCH3

Etuna 2-Butuna

IUPAC Rules for Alkene and Cycloalkene Nomenclature1. 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 groupH2C=CH–CH2– Allyl group

IUPAC Rules for Alkyne Nomenclature1. 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 groupHC≡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-pentena

H2C CCH CH2

CH3

2-metil-1,3-butadiena

H3CC CCH3

2-ButunaH3CC CCH2CH3

C CH

C C

H3C

H CH3

H

trans-2-butena

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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 > alkana

� Baik 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

H

CH

OH2

H3C CH3 -H2O

HC

H3C CH3

Tahap 2. Lepasnya H+

HC

H2C CH3

H

HC

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 CH3CH2X

etil 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 CH2

HCl

(adisi)

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Aturan Markonikov

H3CHC CH2

HClH3CHC 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

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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

OH

2-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+ H2O

H+

CH

OH

CH2CH2CH2CH3H3C