Equation showing formation of a compound

  • Thread starter christinaa_s
  • Start date
  • Tags
    Formation
In summary, when 2-butene reacts with hydrogen chloride gas, only one product is detected, whereas when 1-butene reacts similarily, two products are usually found. This difference can be explained by the fact that the second carbon in 1-butene is chiral.
  • #1
christinaa_s
15
0
Write balanced equations to show the formation of each of the following compunds:
a) butyl propanote
b) propyl methanote


I am doing a horribly constructed independent learning course, they offered absolutly no information on this topic, and then asked me to solve this question..
I really don't even know how to start it.
any help would be much appreciated!
 
Physics news on Phys.org
  • #2
2-butene and 1-butene reactions with HCl

This is another question from my independent learning course:

When 2-butene reacts with hydrogen chloride gas, only one product is detected, whereas when 1-butene reacts similarily, two products are usually found. Explain this.




I don't understand why they would say, "usually" found.

Does the answer have something to do with the fact that the second carbon in 1-butene is chiral?

I was not given very much background information on this topic, only a very brief introduction. The only reason I even know what 'chiral' means, is because my brother, told me that might be one of the reasons.
 
  • #3
christinaa_s said:
Write balanced equations to show the formation of each of the following compunds:
a) butyl propanote
b) propyl methanote

I am doing a horribly constructed independent learning course, they offered absolutly no information on this topic, and then asked me to solve this question..

Hi and welcome to PF christinaa! You've come to the right place. The folks here will be happy to assist you with the difficulties you have in your course.

Both the above compounds are http://en.wikipedia.org/wiki/Ester" reaction. Can you see how you can get each of the products now?

For your second question, have you learned about the rearrangement of Carbocations? If so, can you see how to apply it here?

BTW, what level is your independent learning course and what is the text you are following?
 
Last edited by a moderator:
  • #4
thanx

My independent learning course is at an academic 12 level, and the textbook is Foundations of Chemistry, Second Addition, (Harcourt)

I have not been introduced to the rearrangement of Carbocations in this course. As well we have not been introduced to esterfication, but i will look both these topics up, in order to better understand these two questions.

Thank-you for your help :smile:
 
  • #6
:)

thank-you for the suggestion i will definitely look for that textbook.

For one of the questions I have to draw structural formulas for all the alcohols with the molecular formula of C5H11OH, and give two uses for each type of alcohol, as well as write how they were prepared.

When writing how they were prepared do i just put,
'_____ was prepared by reacting with sulphuric acid solution'
'_____ was prepared by oxidation'
" by hydration...

Also are there only 8 alcohols with the structual formula of (C5H11OH) ?
and, how can I find the uses for the different types of alcohol, i looked then up but i didn't find anything about them. :rolleyes:
 
  • #7
a half-cell balancing question

i really need help with this question. There were about 7 equations that i have to balance by using the half-cell method. I have already finished most of them fairly easily, but these two, i am stuck on. I have been rewriting these 2 equations for the last 7 hours :eek: , but i cannnot succesfully balance them. I don't know what I'm doing wrong..

a) SO3(-2) + MnO4(-) + H(+) <--> Mn(+2) + SO4(-2) + H2O(l)
so far i got this:

its an acid reaction,

=SO3(-2) --> SO4(-2) (was oxidized as 2 electrons were removed)
=SO3(-2) + H2O --> SO4(-2) + 2H(+)
= SO3(-2) + H2O --> S(-2) + 2H(+) + 2e-

= MnO4(-) --> Mn(2+) ( was reduced as 5 electrons were added)
= 5e(-) + MnO4(-) + 8H(+) -->Mn(+2) + 4H2O

b) Cl2(g) + OH(-) <--> Cl(-) + ClO3(-) + H2O(l)

its a basic reaction:

Cl2 --> Cl(-) + ClO(-3) -- (reduced, 2 electrons were added)
OH(-) --> H2O -- (oxidized, 1 electron was removed )
 
  • #8
When writing how they were prepared do i just put,
'_____ was prepared by reacting with sulphuric acid solution'
'_____ was prepared by oxidation'
" by hydration...

It's better if you write the complete reaction, and if possible, the mechanism as well.

a) SO3(-2) + MnO4(-) + H(+) <--> Mn(+2) + SO4(-2) + H2O(l)

From

5e(-) + MnO4(-) + 8H(+) -->Mn(+2) + 4H2O (-I)
and
SO3(-2) + H2O --> SO4(-2) + 2H(+) + 2e(-) (-II)

You have to multiply the equations so that the same number of electrons are involved in both and then add them up

These links should help
http://www.chemguide.co.uk/inorganic/redox/equations.html"
http://www.chemguide.co.uk/inorganic/redox/equations2.html#top"
 
Last edited by a moderator:
  • #9
thanx

Thank-you for your help. The website was ver y helpful. These are the two answers I got. I am not sure about the second one, could you check to see if its right.


a) SO3(-2) + MnO4(-) + H(+) <--> Mn(+2) + SO4(-2) + H2O(l)

=5H2O + 5SO3-2 + 2MnO4(-) + 16H+ <-> 8H2O + 2Mn(+2) + 10H+ + 5S-2


b) Cl2(g) + OH(-) <--> Cl(-) + ClO3(-) + H2O(l)

=2Cl + 4OH(-) + 6HOH(-) --> 6H2O + 2Cl(-) + 2ClO(-3) + 2HOH(-)
 

FAQ: Equation showing formation of a compound

What is a compound?

A compound is a substance made of two or more elements that are chemically bonded together.

What is an equation showing the formation of a compound?

An equation showing the formation of a compound is a representation of the chemical reaction that occurs when two or more elements combine to form a compound.

How do you write an equation showing the formation of a compound?

To write an equation showing the formation of a compound, you must first determine the elements present in the compound and their respective valence numbers. Then, use these numbers to balance the equation by ensuring the number of atoms on both sides of the equation are equal.

Why is it important to balance an equation showing the formation of a compound?

It is important to balance an equation showing the formation of a compound because it follows the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction. Balancing the equation ensures that the number of atoms of each element is the same on both sides of the equation, thus conserving mass.

What information can you learn from an equation showing the formation of a compound?

From an equation showing the formation of a compound, you can learn the elements present in the compound, the ratio in which they combine, and the products formed in the chemical reaction.

Similar threads

Replies
2
Views
14K
Replies
2
Views
3K
Replies
2
Views
3K
Replies
1
Views
1K
Replies
2
Views
4K
Replies
4
Views
5K
Replies
10
Views
4K
Back
Top