Alkenes ( Amino Acids, Protiens And DNA, Aldehydes And Ketones )

Question 1

Dodecane ( C12H26 ) is a hydrocarbon found in the naphtha fraction of crude oil.
Dodecane can be used as a starting material to produce a wide variety of useful products.
The scheme below shows how one such product, polymer Y, can be produced from dodecane.

C12H26 - > C2H4 + C4H8 + x

nC4H6 < - > - ( CH( CH3 )CH( CH3 ) ) -n
Polymer Y

( First arrow is labelled “ Reaction 1 “ )

( The second arrow is labelled “ Reaction 2 “ )

Name the homologous series that both C2H4 and C4H8 belong to.
Draw a functional group isomer of C4H8 that does not belong to this homologous series.

Answer 1

Name:

• Alkenes

Functional group isomer:

• CH2CH2CH2CH2 ( Carbons are connected to eachother in a square )
• ( Cyclobutane )

Question 2

Identify compound X.

( Dodecane ( C12H26 ) is a hydrocarbon found in the naphtha fraction of crude oil.
Dodecane can be used as a starting material to produce a wide variety of useful products.
The scheme below shows how one such product, polymer Y, can be produced from dodecane.

C12H26 - > C2H4 + C4H8 + x

nC4H6 < - > - ( CH( CH3 )CH( CH3 ) ) -n
Polymer Y

( First arrow is labelled “ Reaction 1 “ )

( The second arrow is labelled “ Reaction 2 “ ) )

Answer 2

Compound x:

• C6H14

Question 3

Name polymer Y.

( Dodecane ( C12H26 ) is a hydrocarbon found in the naphtha fraction of crude oil.
Dodecane can be used as a starting material to produce a wide variety of useful products.
The scheme below shows how one such product, polymer Y, can be produced from dodecane.

C12H26 - > C2H4 + C4H8 + x

nC4H6 < - > - ( CH( CH3 )CH( CH3 ) ) -n
Polymer Y

( First arrow is labelled “ Reaction 1 “ )

( The second arrow is labelled “ Reaction 2 “ ) )

Answer 3

Name of polymer Y:

• Poly( but-2-ene )

Question 4

Reaction 1 is an example of thermal cracking and is carried out at a temperature of 750 °C.
State one other reaction condition needed.

( Dodecane ( C12H26 ) is a hydrocarbon found in the naphtha fraction of crude oil.
Dodecane can be used as a starting material to produce a wide variety of useful products.
The scheme below shows how one such product, polymer Y, can be produced from dodecane.

C12H26 - > C2H4 + C4H8 + x

nC4H6 < - > - ( CH( CH3 )CH( CH3 ) ) -n
Polymer Y

( First arrow is labelled “ Reaction 1 “ )

( The second arrow is labelled “ Reaction 2 “ ) )

Answer 4

Condition needed for thermal cracking:

• High pressure

Question 5

Reaction 2 is exothermic.
A typical compromise temperature of 200°C is used industrially for this reaction.
Explain the effect of a change of temperature on both the position of equilibrium and the rate of reaction, and justify why a compromise temperature is used industrially.

( Dodecane ( C12H26 ) is a hydrocarbon found in the naphtha fraction of crude oil.
Dodecane can be used as a starting material to produce a wide variety of useful products.
The scheme below shows how one such product, polymer Y, can be produced from dodecane.

C12H26 - > C2H4 + C4H8 + x

nC4H6 < - > - ( CH( CH3 )CH( CH3 ) ) -n
Polymer Y

( First arrow is labelled “ Reaction 1 “ )

( The second arrow is labelled “ Reaction 2 “ ) )

Answer 5

The effect of changing the temperature:

• Reaction is exothermic
• Increasing temperature favours the endothermic reaction
• Therefore equilibrium will shift to the left
• Decreasing the yield of product
• At higher temperatures, more molecules have higher kinetic energy
• More molecules have energy higher than the activation energy
• So rate of reaction increases
• The compromise temperature is used to keep cost low and to produce a suitable yield of product

Question 6

Repeating units of two polymers, P and Q, are shown in the figure below.

P = - CH( CH3 )CCl( CH3 ) -

Q = - OCH( CH3 )CH( CH3 )OCOC( CH3 )2CH2CO -

Draw the structure of the monomer used to form polymer P.
Name the type of polymerisation involved.

Answer 6

Monomer used to form polymer P:

CH3CH = CClCH3

Type of polymerisation:

• Addition polymerisation

Question 7

Draw the structures of two compounds that react together to form polymer Q.

( Repeating units of two polymers, P and Q, are shown in the figure below.

P = - CH( CH3 )CCl( CH3 ) -

Q = - OCH( CH3 )CH( CH3 )OCOC( CH3 )2CH2CO - )

Answer 7

Sturcture of Compound 1:

• HOCH( CH3 )CH( CH3 )OH

Structure of Compoind 2:

• HOOCC( CH3 )2CH2COOH

Question 8

Suggest an environmental advantage of polymer Q over polymer P.
Justify your answer.

( Repeating units of two polymers, P and Q, are shown in the figure below.

P = - CH( CH3 )CCl( CH3 ) -

Q = - OCH( CH3 )CH( CH3 )OCOC( CH3 )2CH2CO - )

Answer 8

Advantage:

• Q is Biodegrable

Justification:

• Q has a polar C = O
• Therefore Q can be attacked by nucleophiles ( leading to its breakdown )

Question 9

The amide or peptide link is found in synthetic polyamides and also in naturally occurring proteins.

Draw the repeating unit of the polyamide formed by the reaction of
propanedioic acid with hexane-1,6-diamine.

Answer 9

Structure of reapeating unit:

• -COCH2CONHCH2CH2CH2CH2CH2CH2NH-

Question 10

In terms of the intermolecular forces between the polymer chains, explain why polyamides can be made into fibres suitable for use in sewing and weaving, whereas polyalkenes usually produce fibres that are too weak for this purpose.

( The amide or peptide link is found in synthetic polyamides and also in naturally occurring proteins. )

Answer 10

Polyamides can be made into fibres suitable for use in sewing and weaving, whereas polyalkenes usually produce fibres that are too weak for this purpose because:

• Polyamindes contain hydrogen bonding
• Polyaklenes only have van der waals forces
• Hydrogen bonding is stronger than van der waals forces

Question 11

Name and outline a mechanism for the reaction of CH3CH2COCl with CH3NH2

Answer 11

Name of mechanism:

• Nucleophilic addition elimination

Mechanism:

• ( NH2, with lone pair of electrons, attacks the delta positive carbon attached to the O )
• ( Arrows are drawn from the lone pair on NH2 to the carbon, then an arrow from the double bond in C = O to the oxygen )
• ( The oxygen now has a lone pair of electrons and a negative charge )
• ( The newly bound CH3NH2 now has a postive symbol on the N atom )
• ( The lone pair of electrons on the O reform the double bond )
• ( As this happen, a H from the CH3NH2^+ is lost from the N atom, illustated by an arrow from the N - H bond to the positive N atom )
• ( Also in this step, the Cl is lost illustated by an arrow from the C - Cl bond to the Cl atom )
• Show the formed product

Question 12

Give the name of the product containing an amide linkage that is formed in the reaction in part ( b ) ( i ).

( The amide or peptide link is found in synthetic polyamides and also in naturally occurring proteins. )

( CH3CH2CONHCH3 )

Answer 12

Name of the amide:

• N-methylpropanamide

Question 13

2c