Alcohols to alkenes Flashcards

(28 cards)

1
Q

What occurs in elimination reactions?

A

2 sigma bonds are broken and a pi bond forms

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2
Q

Propan-2-ole elimination reactions explained

A

1) sigma bond between the C—OH bond breaks
2) carbon adjacent to it (left carbon) hydrogen breaks H—C
So One H is gone and another OH is gone forming water as well
3) a pi bond forms
forming the alkene propene and water

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3
Q

Whats the trend between elimination reactions in alcohols?

A

it always forms an alkene and water

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4
Q

Another example

A

OH bond breaks pushing C—H towards the side

the second carbon will lose its hydrogen too

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5
Q

Whats dehydration reaction defined as

A

when a molecule loses an atom from its molecule

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6
Q

To get an elimination reaction or dehydration reactions what do we need

A

state• concentrated acid catalyst
Reagent being = (concentrated sulfuric acid H2SO4) or (concentrated phosphoric acid H3PO4)

+ Heat the mixture to increase the rate of reaction

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7
Q

What is the name of this type of reaction?

A

electrophilic
as it involves breaking 1 pi bond and forming 2 sigma bonds so it’s an additional reaction as HBr acts as an electrophile

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8
Q

State a reagent for this dehydration reaction

A

H2SO4 or H3PO4

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9
Q

State the role of this reagent for the dehydration reaction

A

it’s a catalyst

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10
Q

The conditions needed for a dehydration reaction or elimination reaction is

A

concentrated acid catalyst and heat

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11
Q

Name the products of the dehydration reaction of propan-1-ol

A

propene and water

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12
Q

What does this arrow tell us between the reaction of H2SO4 and Propen-2-one

A

tells us that hydrogen hydrogen bond is formed

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13
Q

What does this arrow tell us

A

tells us a hydrogen oxygen bond is broken

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14
Q

What’s happening here?

A

the fourth molecule OH molecule on the OH2 gained a H so it gains a + charge

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15
Q

Whats does the arrow tell us here

A

this arrow tells us that the +OH2–C bond breaks so we form a carbocation
and the +OH2 broken forms water

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16
Q

What’s happening in the final step?

A

First big arrow tells us the Oxygen is forming a bond with the Hydrogen
and second small arrow tells us a pi bond is formed which forms this

17
Q

Because sulfuric acid is a reagent at the beginning of the reaction and a product at the end, it’s acting as a

18
Q

another example

A
  1. OH lone pair goes to H & bond breaks oxygen bond
  2. OH2+ bond breaks forming water
  3. Carbocation occurs
  4. Left element H attracts the lone pair of Oxygen on sulfuric acid and bond breaks
    whilst the C—H bond breaks to make a Pi bond for C—C
19
Q

what’s the mechanism for ethanol to ethene?

20
Q

We can replace H2SO4 by a H+ like this

A

This is the simplified version

so the mechanism would look like

21
Q

What are these molecules examples of and what are they called?

A

1st molecule is called Pent-1-ene
and second molecule is called Pent-2-ene
both examples of structural isomers

22
Q

Pent-2-ene has how many substituents

A

it has 2 substituents so it’s a z-pent-2-ene and e-pent-2-ene

23
Q

When hex-3-ene undergoes elimination it can produce…

A

it forms hex-2-ene, hex-3-ene

which has both 2 isomers as shown

24
Q

how do we write this elimination reaction

A

CH3CHOHCH3 —> C3H6 + H2O

if there’s the double bond only goes one way so 1 for both we write the molecular formula

25
write the elimination reaction of ethanol to ethene
CH3CHOH —> C2H4 + H2O
26
how do we write the equation for this reaction?
CH3CHOHCH2CH2CH3 —> CH3CH=CHCH2CH3 + H2O
27
When this alcohol undergoes elimination what will form?
First dot near OH = 1 molecule | second dot near upper tail = second molecule
28
How did the third molecule form?
tertiary structures are more stable than secondary so the secondary carbocation intermediate may undergo rearrangement and carbocation rearranges to its atom to form a tertiary