CH203 17c Alkene Probs 2

8/17/2019 CH203 17c Alkene Probs 2

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Addition to carbon–carbon double bonds • Problems © Bruno I. Rubio 1

CH 203

O R G A N I C C H E M I S T R Y I

Addition to carbon–carbon double bonds: Problems

(1) Predict the organic product(s) of the reaction of isobutylene (atright) with each of the following reagents. Assume that no carboca-

tion rearrangements occur.

(a) H2, Pt (b) Br2

(c) HBr (d) H2O, H2SO4, 100 °C

(e) BH3 then H2O2, NaOH, H2O (f) FCO3H

(g) Cl2, H2O (h) OsO4 then NaHSO3

(i) O3 then Zn, CH3CO2H (j) O3 then H2O2

(k) Cl2, CH3CH2OH

(2) Predict the organic product(s) of the reaction of 2-

methylpent-2-ene (at right) with each of the reagents in Prob-

lem (1). Indicate which reactions afford a mixture of enanti-

omers. Assume that no carbocation rearrangements occur.

(3) Predict the organic product(s) of the reaction of 1-

methylcyclopentene (at right) with each of the reagents in Prob-

lem (1). Clearly show the stereochemistry of the product(s) where appropri-

ate. Indicate which reactions afford a mixture of enantiomers. Assume that no

carbocation rearrangements occur.

(4) Predict the organic product of the reaction of (S )-6-

methylbicyclo[4.4.0]dec-1-ene (at right) with each of the rea-

gents in Problem (1). Clearly show the stereochemistry of the

product where appropriate. Assume that no carbocation rearrange-

ments occur.

(5) Hydrogen occurs in the three isotopic forms protium (1H or simply H; this

is plain ol’ hydrogen), deuterium (2H or D; about 0.05% of all hydrogen atoms

are D atoms) and tritium (3H or T; the only radioactive isotope of hydrogen).

The isotopes differ in the number of neutrons in the nucleus. Except for oc-

casional differences in the rate of reaction, the three hydrogen isotopes re-

act essentially identically. Predict the organic product(s) of the reaction

of 1-methylcyclopentene (see problem (3)) with each of the following deute-

rium-containing reagents. Clearly show the stereochemistry of the product(s).

CH3

H3C

CH3

CH3

CH3H3C

CH3


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If more than one product results, indicate how each is related to the

other(s).

(a) D2, Pt (b) DCl (c) BD3 then H2O2, NaOH, H2O

(6) Predict the products of the reaction of IBr with (S )-6-

methylbicyclo[4.4.0]dec-1-ene (see Problem (4)), clearly showing the stereo-

chemistry of the product. Write a detailed mechanism of the reaction, using

curved arrows to show the flow of electrons.

(7) Circle the alkenes that are likely to undergo a carbocation rearrangement

upon addition of HCl. In those cases where rearrangement is likely, draw the

structure of the final (rearranged) organic product.

H3C

CH3

CH3H3C

CH3

CH3

H3C CH3

(8) Rank these alkenes in order of their rate of reaction with HBr (1 = fast-

est).

CH3

OCH3

CH3

NCH3H3C

CH3CH3

CH3

(9) Each of these reactions affords only one of the two products shown below.

Circle the product that forms and explain the reasoning for your selection.

(a)

H3

CCF

3

HCl H3

CCF

3H3C

CF3

or

Cl

Cl

(b)

H3

COCH

3

HCl H3

COCH

3H3C

OCH3

or

Cl

Cl

(10) Write detailed electron-pushing mechanisms of these reactions, using

curved arrows to show the flow of electrons.


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(a)

CH3 CH3

H3

C Cl

HCl

(b)CH

3

H3C CH3

HClH3C

Cl

(c)

NH Br2

N

Br

+ HBr

(d)

CH3

CH3

OH

Cl CH3

CH3

+ HCl+ H2O

(e)

CH3

H3C

I

O

O

CH3

H3C

O

HO + I2 + HI

The use of hydrogenation and ozonolysis in structure determination

(11) A pine-scented hydrocarbon (C15H24) isolated from a cypress tree reacts

with excess H2 over a Pt catalyst to afford the bicyclic alkane shown below:

C15H24

excess

H2

Pt

H3C

CH3

CH3

CH3

When C15H24 undergoes reaction with excess O3 followed by work-up with zinc

and acid, the following products are obtained:


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H3C

CH3

CHO

CH3O

OC15H24

(1) excess O3

(2) Zn CH3CO2H

+ CH2O

Suggest a reasonable structure for C15H24.

(12a) Draw six additional, reasonable resonance struc-

tures of the molecule at right.

(12b) By inspecting the resonance structures you drew,

predict which atoms are likely to undergo attack by a nu-

cleophile.

CN

OCH3


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ANSWERS

(1a)H

3C CH

3

CH3

(1b)

CH3

CH3

Br

Br

(1c)H3C

CH3

CH3

Br (1d)H3C

CH3

CH3

OH

(1e)CH

3

CH3

HO

(1f)

CH3

CH3

O

(1g)

CH3

CH3

OH

Cl

(1h)

CH3

CH3

OH

HO

(1i)

CH3

CH3

O

+

H

H O

(1j)

CH3

CH3O

(1k)

CH3

CH3

OCH2CH

3

Cl

(2a)H

3C CH

3

CH3

(2b)

H3CCH3

CH3

BrBr

(±)

(2c)

H3CCH3

CH3

Br (2d)

H3CCH3

CH3

HO

(2e)

H3C CH3

CH3

OH

(±)

(2f)

O

H3C

H3CCH3

(±)

(2g)

H3CCH3

CH3

HOCl

(±)

(2h)

H3CCH3

CH3

HOOH

(±)


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(4e)

CH3

HOH (4f)

CH3

O

(4g)

CH3

OHCl (4h)

CH3

OHOH

(4i) O

CH3H

O

(4j) O

CH3HO

O

(4k)

CH3

OCH2CH3Cl

(5a)

D

CH3

D

(±)

(5b)

Cl

CH3

D

(±)

Cl

CH3

D

(±)

+

(5c)

D

CH3

OH

(±)

(6) The iodine–chlorine bond is polarized away from iodine because iodine is

less electronegative than chlorine. Thus, the iodine end of ICl is partially

positive and undergoes attack the the alkene:

CH3

ClI

CH3

I Cl

!+ !"

– Cl –

CH3

I

Cl


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(7)

H3C

CH3

CH3H3C

CH3

CH3

H3C CH3

H3C

CH3

CH3Cl

CH3

CH3

CH3

Cl

Cl

CH3

HCl HCl

HCl

(8) The decisive factor is the relative stability of the carbocation formed

when the carbon–carbon double bond is protonated by HBr.

CH3

OCH3

CH3

NCH3H3C

CH3CH3

CH3

2 4 5 1 3

(9a)

H3CCF3

HCl H3CCF

3H

3C

CF3

or

Cl

Cl2

3

The –CF3 group is electron-withdrawing by induction and would destabilize a

carbocation at C2; hence, protonation occurs at C2 and the carbocation formed

at C3 is captured by Cl – .

(9b)

H3

COCH

3

HCl H3

C

OCH3H3C

OCH3

or

Cl

Cl1

2


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The –OCH3 group is electron-donating by resonance and stabilizes the carboca-

tion at C1.

(10a)

CH3 H Cl

– Cl – CH3

CH3

CH3

CH3

=

CH3

CH3

Cl –

product

(10b)

CH3

H3C CH3

H Cl

H3C CH3

– Cl –

Cl –

product

H3C CH3

=

H3C

(10c)

Br Br

– Br –

NH

NH

BrN

H Br

=N

H

Br

Br –

products


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(10d)

Cl CH3

CH3

CH3

CH3

CH3

CH3

– Cl

CH3

CH3

OH2

CH3

CH3

O

H

HCl

– HCl

product

(10e)

CH3

H3C I

O

O

CH3

H3CO

HO

I I

– I

CH3CH3

O

HO

I

H

I

– HI

product

(11)

H3C

CH3

CH3


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(12a)

CN

OCH3

CN

OCH3

CN

OCH3

C

N

OCH3

C

N

OCH3

CN

OCH3 C

N

OCH3

(12b)

All of the carbon atoms that bear a positive charge may be attacked by a nu-

cleophile. Oxygen, though positive, cannot be attacked by a nucleophile be-

cause, if that were to happen, the octet of the oxygen would be exceeded.

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CH203 17c Alkene Probs 2