2-methylbutan-1-ol is heated with an excess of concentrated sulphuric acid (H2SO4) in order to cause an elimination reaction:
H3C-CH2-CH(CH3)-CH2OH + H2SO4 ==> H3C-CH2-C(CH3)=CH2
2-methylbutan-1-ol + concn H2SO4 ==> 2-methylbut-1-ene + water
1) Draw all of the above as displayed formula.
2) What is the product of this reaction?
2) Provide a ‘curly arrow’ mechanism (in displayed formula) for
the formation of the product supported with an explanation.
1)
----H--H--CH3--H -----------------------------O ----------------H ----- CH2CH3
-----I----I----I----I -------------------------------II -------------------\ -------/
H - C - C - C - C - O - H -- +----- H - O - S -O -H ==> ---- C = C ------ + how
-----I-----I----I----I ------------------------------ II ------------------/ ------ \ --- do I draw
----H----H---H---H -----------------------------O --------------- H ------- CH3 -- water
2) Is the product of the reaction:
2-methylbut-1-ene + water
Why does it ask for the product when there are two products. Why does it not ask what are the two products or what are the products?
As the answer for what is the product so I write:
2-methylbut-1-ene + water
3) ?
Why not ask for productS instead of product. The only person that can ask that is the author of the problem.
My attempt at H2O follows:
-----------H
------------|
------------O-H
The length of the vertical O-H bond and the length of the horizontal O-H should be the same but I have no control over that.
3. Can't do curly arrows or any other kind of arrows for mechanisms. Sorry about that.
2-methylbut-1-ene
Option 1:
-----H---H----------H
-----I-----I----------/
H - C - C - C = C
-----I----I----I------\
----H---H---CH3- H
Is the displayed formula for 2-methylbut-1-ene above right?
Option 2
----H---------H---H
-----I----------I----I
H - C = C - C - C - H
-----I-----I-----I----I
----H--CH3--H---H
Option 3
H-------H--------H
---\------I---------I
--- C = C - C - C - H
---/------I-----I----I
H-------H-CH3--H
These two structures above came up online.
Which ones are right?
3) Provide a ‘curly arrow’ mechanism (in displayed formula) for
the formation of the product supported with an explanation.
Do I copy the displayed formula I did for part 1 and then add arrows?
Where would I add the arrows if this is the case?
What would the explanation be?
------------H
------------|
------------O-H
Is this the exact way water must always be drawn when asked for its displayed formula?
3)
Step 1:
----H--H--CH3--H -----------------------------O
-----I----I----I----I -------------------------------II
H - C - C - C - C - O - H -- +----- H - O - S - O -H
-----I-----I----I----I -- .. ------------------------- II
----H----H---H---H --I --------------------------O
------------------------ --> H+
==>
-----H---H----------H
-----I-----I----------/ ---------------- H
H - C - C - C = C -------- + ----- I
-----I----I----I------\ -----------------O - H
----H---H---CH3- H --------------
=========================================>
----H--H--CH3--H ------------------------------O
-----I----I----I----I --------------------------------II
H - C - C - C - C - O - H -- +----- H - O - S - O -H
-----I-----I----I----I -- + ------------------------- II
----H----H---H---H -/ \ --------------------------O
-----------------------H H
==>
-----H---H----------H
-----I-----I----------/ ---------------- H
H - C - C - C = C -------- + ----- I
-----I----I----I------\ -----------------O - H
----H---H---CH3- H --------------
Step 2:
----H--H--CH3--H _ ----------------------------O
-----I----I----I----I -/--I-----------------------------II
H - C - C - C - C^ - O - H -- +----- H - O - S - O -H
-----I-----I----I----I -- + -------------------------- II
----H----H---H---H -/ \ ---------------------------O
-----------------------H H
==>
-----H---H----------H
-----I-----I----------/ ---------------- H
H - C - C - C = C -------- + ----- I
-----I----I----I------\ -----------------O - H
----H---H---CH3- H --------------
=========================================>
----H--H--CH3--H _ ----------------------------O ------------H
-----I----I----I----I -/--I-----------------------------II ------------I
H - C - C - C - C^ - O - H -- +----- H - O - S - O -H --- O -- H
-----I-----I----I----I -------------------------------- II
----H----H---H---H -------------------------------O
-----------------------
==>
-----H---H----------H
-----I-----I----------/ ---------------- H
H - C - C - C = C -------- + ----- I
-----I----I----I------\ -----------------O - H
----H---H---CH3- H --------------
=========================================>
Step 3
----H--H--CH3--H _ --------------------------O ------------H
-----I----I----I----I -/--I---------------------------II ------------I
H - C - C - C - C^ - O - H -- +----- H - O - S - O -H --- O -- H
-----I-----I----I----I +----------------------------- II
----H----H---H---H -----------------------------O
-----------------------
==>
-----H---H----------H
-----I-----I----------/ ---------------- H
H - C - C - C = C -------- + ----- I
-----I----I----I------\ -----------------O - H
----H---H---CH3- H --------------
=========================================>
----H--H--CH3--H ----------------------------O ----
-----I----I----I----I -------------------------------II -------
H - C - C - C - C - O - H -- +----- H - O - S - O -H ----- CH2 = CH2 --- H+
-----I-----I----I----I------------------------------- II
----H----H---H---H -----------------------------O
-----------------------
==>
-----H---H----------H
-----I-----I----------/ ---------------- H
H - C - C - C = C -------- + ----- I
-----I----I----I------\ -----------------O - H
----H---H---CH3- H
I got confused for question 3.
3) Provide a ‘curly arrow’ mechanism (in displayed formula) for
the formation of the product supported with an explanation.
-----------------CH3---------------------------------------------CH3
-------------------I-------------------H2SO4 ---------------------I
CH3 - CH2 - CH - CH2 - OH ======> CH3 - CH2 - C = CH2 + H2O
I saw online that someone put the H2SO4 in a different formula on top of the arrow. I did this for the formula that was written here too. Why is the H2SO4 put over the arrow instead of added before the arrow?
Is this what the displayed formula should look like for this question? Or was I supposed to draw everything out as the displayed formula was asked in the question?
Do I write out the sulphuric acid like below:
------O
-------II--------
OH - S - OH
-------II--------
-------O
3)
Step 1
-----------------CH3-------------------------------------CH3
-------------------I-------------------------------------------I
CH3 - CH2 - CH - CH2 - OH ==> CH3 - CH2 - CH - CH2 - +OH
--------------------------------- .. --------------------------------------------I
----------------------------------^ --------------------------------------------H
----------------------------------I
----------------------------------H+
=================================================>
Step 2
-----------------CH3 -^-____ ------------------------------CH3
-------------------I -----I ------I -------------------------------I ----+
CH3 - CH2 - CH - CH2 - +OH ==> CH3 - CH2 - CH - CH2
-------------------------------------I
-------------------------------------H
-------------------------H
--------------------------I
-------------------------O - H
=================================================>
Step 3
-----------------CH3__/
-------------------I -/-+^--------- ==> CH3 - CH2 - C (CH3) = CH2
CH3 - CH2 - CH - CH2
----------------------------------------------------------------H+
Is this correct?
Option 1 is correct for 2-methylbut-1-ene
Option 2 would be correct if 1 H atom was removed from C1. As it is C1 is a five valent C and that's not allowed.
Option 3 has a trivalent c arbon on C3 so you need to add another H to the C3 carbon. This structure, with C3 corrected is 3-methylbut-1-ene
I can't help with the mechanism and drawing of arrows/ curves/etc. I tried to find something on the web I could copy. If ound plenty of examples but couldn't copy and/or copy and paste to this site. I suggest you go to Google and type in something like "examples showing use of curved arrows to show the mechanism of organic chemistry reactions". The way you show it above in step 1, step 3, etc is not the way it's done in texts. The steps are there BUT they have curved arrows to show you where electrons are being moved from and where they are being moved to. Get a chemistry text and it will be over flowing with examples.
To be very honest about it I've never seen the term "displayed form" used and I'm not sure what is meant by that. What you have been drawing and calling displayed form I have always called structural formula. The condensed formula for butane, for example, is CH3CH2CH2CH3 and the final one is the empirical formulaof C4H10.
Finally no water can be shown in a number of ways. It can be written as H2O or HOH, or the Lewis structure which I can draw here with the dots, etc, and the way I drew it above is not correct technically. That bond angle between the two H atoms is 90 degrees in my drawing but it should be about 105 degrees. If you played with the \ and / and | it might look better and closer to 105 degrees but the only way to do it right is to use a program like CHEMDRAW, or a similar one. Chemdraw also does a great job of drawing the structural formula AND plops in the curved arrows when directed to do so.
The question means structural formula when it says displayed formula.
Is my last post before your post right?
Ignore the steps and arrows separating the steps.
There is only a ^ arrow and not a down arrow so imagine that those are facing the opposite way.
This is an explanation I found online for when 2-methylpropan-1-ol is heated with concentrated sulphuric acid for an elimination reaction:
Elimination in alcohols
The dehydration is catalysed by strong acid, usually sulfuric or phosphoric acid, which promotes the protonation of the hydroxyl oxygen atom. The mechanism proceeds in two main steps, with different intermediates between primary and tertiary alcohols:
1. Protonation
2. Elimination
Mechanism step 1
(first step will be put here)
Protonation of the hydroxyl group oxygen, forming a positive oxonium ion
Elimination in alcohols
Mechanism - step 2
(second step will be put here)
The bond holding the oxygen to the carbon atom then breaks with the electron pair going onto the oxygen to cancel out the charge and making water, a good leaving group.
This leaves a positive charge on the carbon atom that previously held the oxygen.
This ion would be unstable for primary carbocations and so only occurs with tertiary alcohols.
Elimination in alcohols
Mechanism - step 3
(third step would be put here)
The bond holding the hydrogen atom to the adjacent carbon atom then breaks and the electron pair go to the pi orbital between the two carbon atoms. (This happens in concert with the previous step in primary alcohols to avoid the formation of the primary carbocation).
The positive charge on the carbon atom is then cancelled out and a positive hydrogen ion is released restoring the acidity of the solution.
Hence the dehydration is catalysed by hydrogen ions.
Does this explanation make sense for part 3) of my question? Should this be written out?
It said online:
The displayed formula and structural formula are fairly SIMILAR. The displayed formula represents the molecule by showing all of the atoms, and all of the bonds between those atoms. Bonds are represented by lines, with the number of lines representing the strength of the bond.
A structural formula shows how the various atoms are bonded. There are various ways of drawing this and you will need to be familiar with all of them. A displayed formula shows ALL the bonds in the molecule as individual lines. You need to remember that each line represents a pair of shared electrons.