Concentrated sulphuric Dehydration acid of (H alcohols 2SO4) or phosphoric acid (H3PO4) : as acidic alcohols catalysts and dehydrating agents. C
C
H
OH
H2 SO4 (conc.)
C
C
+
H2 O
The major product is the MOST STABLE ALKENE, which has greater number of alkyl group attach to C=C is follow SAYTZEFF’S RULE.
Mechanism of dehydration of alcohols
3 step of mechanism
• Protonati on of alcohols
Step 1
Step 2
• Formation of carbocation
• Formatio n of alkenes
Step 3
Step 1 : Protonation of alcohols H H has form A bond + as aOresult from H C C CH CH : H + is 3 2 What 3 protonation Why it has “+” :OH H .. protonation? (nucleophile
(nucleophile charge? reacts with acid, attacking atom OxygenHhas loss + from H 0 3 H of Electron its pair H + + move to electron (make a H or H 0 3 Nucleophile (lone pair at + H C C CH CH O HH : Oxygen (look bond with 3atom) 2 3 .. O attack + at arrow atom) and it also : OH electrophile movement) (H atom has +1 formal H acid) form charge
Step 2 : Formation of carbocation How to neutralize the H H “+” charge? + H C C CH CH Move the electron H C C CH CH form 3 2 3 2 3 3 + :OH Carbon to Oxygen (look carbocation at arrow movement). H This will results a H carbocation formation. 0 3 Redraw the + :O H carbocati previous .. on structure from step 1
Step 3 : Formation of alkenes H H H H C C+ C CH
3
H
H
b
a OR
In order to form an alkenes, all the bond surrounded carbocation has Lets look at these twopossibilities possibilitiesto form a double 1.Root a a bond 2.Root b (Look at arrow movement)
Step 3 : Formation of alkenes Root b
alkene a
Root a
H H H H H H C C+ C CH + : O H C .. 3 H C H H H2O alkene + (determine b Hfrom step 1) H asH a : Onucleophile H C C
..
H
3
+
CH2CH3
H C +
H + :O H
CH 3
H
How to determine major and minor product? H H H H Saytzeff’s rule H H + An elimination the : O H gives H C C C CH +usually C C .. 3 most stable alkene product, b H highly C CH a the most H commonly H 3 3 + + substituted alkene.major H H
H
:O H ..
H
C H
+ + :O C CH2CH3 H
minor
H
Rearrangement During Dehydration of Alcohol
Example 1: CH H HC 3
C
3
C
CH OH
CH
H SO (conc.) 2
4
3
3
You might predict the structure beside but the answer is wrong. But WHY?
CH H HC 3
C
3
CH 3
C
CH 2
Rearrangement During Dehydration of Alcohol
Example 1: CH H HC 3
C
3
C
CH OH
CH 3
H SO (conc.) 2
4
3
This is because CH CH3 3 rearrangement H C C CH CH + actual H 3C C occurs. C C H 3The 2 3 is…. C Hproduct CH 3 3 (major product) (minor product)
Step 1 : Protonation of alcohols CH H HC 3
C
3
C
CH +
CH :OH ..
3
H + :O H H
3
CH H H
:O ..
+ H
HC 3
C
3
C CH 3 + CH :OH 3
H
Step 2 : Formation of carbocation The rearrangement CH H through the occurs CH H 3 migration of an alkyl 3 H Cgroup C C CH from theH C C C+ CH (methyl) 2 1 3 3 4 + atom adjacent to 3 0 3 0 carbon 0 CH :OH CH the one 3 with the positive 2 3 H charge. carbocation Rearrangement 0
1,2-methy shift
Because group CHa H 3 the one migrates from H Cto the C next, C CH carbon this 3 + kind of3 rearrangement is 3 carbocation CH often called a 1,23 shift. More stable 0
Less stable
Step 3 : Formation of alkenes
H
(a)
+
H CH H H- C H
C +
3
C
CH HC 3
:O H ..
(b)
CH
3
Major product
3
CH
3
+
C
CH
CH
3 C CH
H :O H ..
HC 2
C CH
3 CH CH 3
3
Minor product
3
Try this! CH H 3 H C C C CH 3 3 H OH
H2SO4 (conc)
+
CH 3 H C C CH CH 3 3
CH 3 H C C CH CH 2 2 3