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Topic: Carbonyl Compounds and some Thermodynamics  (Read 4062 times)

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Offline AmbitiousStudent

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Carbonyl Compounds and some Thermodynamics
« on: November 03, 2013, 03:16:02 AM »
I am doing one task and I need some help to see if I am on the right direction. Please post the answers so I had something to compare with but also why it will be like this.

Part 1 Draw a Lewis structure for acyl chloride R-C(=O)-Cl and state its 3D-structure around the central atom (C). R is an acyl substituent.

I guess it just will be standard AX3 and be trigonal planar according to the VSEPR theory. Correct?

Part 2 In a substituent reaction will the leaving group (L=Cl- ) be substitute by a new substituent. Is the substituent in the carbonyl compound nucleophile or electrophile? Argue according to your answer in part 1.

Nucleophile. Cl- will be affinietied to the nuclei. Will be seen in the drawing above since Cl will have 7 electrons. Correct?

Are L=HCOO- (methanoate) better or worse leaving group compared to Cl- ?. Explain why it is better or worse.

A good leaving group is a weak base. So check up the pKa value and HCl have -9 and HCOOH have 3,75. Correct?

Part 3 Acryl chloride HC(=O)Cl can be produced by a substitution of methanoate HCOOH even if OH- is worse as a leaving group compared to Cl- . The reason why is that you can use thionyl chloride ClS(=O)Cl as a substitution reagent in a reaction where it also will HCl and SO2 will be arise. Write a balanced reaction formula?

Is this one correct HCOOH + ClSOCl = HCOCl + SO2 + HCl ?

Part 4 According to the the table, the driving force is because HCl and SO2 is gases Check this up with calculate the reaction enthalpy and reaction entropy at 298K. If you need more data will you have to check it up in a table work

ClS(O)Cl (aq): Delta Hf -785 (kJ/mol) Sm 220 (J/K/mol)
HC(O)Cl(aq) :Delta Hf -376.56 (kJ/mol) Sm 246.52 (J/K/mol)

Offline zsinger

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Re: Carbonyl Compounds and some Thermodynamics
« Reply #1 on: November 03, 2013, 01:33:54 PM »
So whats your question?  That sounds like a carbon copy take home test that you posted on here for us to answer?
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"

Offline AmbitiousStudent

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Re: Carbonyl Compounds and some Thermodynamics
« Reply #2 on: November 03, 2013, 02:48:44 PM »
My question is that I wonder if I am on the right track. I prefer if someone can post the right solutions and if that is too much I would really appreciate if I can get some hints. It is some kind of a home work but at the same time not. It is a task from a book with no answers. If you are not allowed to post these kind of questions in this section would I be glad if you can tell me where I can post it.

Offline spirochete

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Re: Carbonyl Compounds and some Thermodynamics
« Reply #3 on: November 03, 2013, 02:53:09 PM »
So whats your question?  That sounds like a carbon copy take home test that you posted on here for us to answer?

That could certainly be possible but it's very hard to tell on the internet. He did actually make an effort to answer all the questions, which people are required to do.

I'd say all your answers look reasonable except I didn't have time to look at the last one. Also I didn't count up the atoms for you to double check that everything is balanced on part 3, but those are definitely the side products. That's why pyridine (a weak base) is a common additive for rxns like that.

Also I don't think you finished explaining part 2 but you are on the right track. At least you didn't quite give the answer.

Edit: just saw his response. A mod can let me know how we handle people being honest like that. In my view he could just as easily have a friend who happened to be a chem major to double check his answers, but I'm not 100% sure of forum policy.

Offline AmbitiousStudent

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Re: Carbonyl Compounds and some Thermodynamics
« Reply #4 on: November 04, 2013, 05:40:56 PM »
Someone else who had some answers or some tips how I should continue? And I am sorry I do not have any friend with a Chem major who could check my answers.

Offline spirochete

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Re: Carbonyl Compounds and some Thermodynamics
« Reply #5 on: November 04, 2013, 05:44:00 PM »
Someone else who had some answers or some tips how I should continue? And I am sorry I do not have any friend with a Chem major who could check my answers.

In my last post I said they all look good except you need to completely answer one, although your line of logic was correct so far.

The only one I didn't look at was the last one.

Offline AmbitiousStudent

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Re: Carbonyl Compounds and some Thermodynamics
« Reply #6 on: November 05, 2013, 05:51:56 PM »
Part 4 According to the the table, the driving force is because HCl and SO2 is gases Check this up with calculate the reaction enthalpy and reaction entropy at 298K. If you need more data will you have to check it up in a table work

HCOOH+ClSOCl -> HCOCl+HCl+SO2

ClS(O)Cl (aq): Delta Hf -785 (kJ/mol) Sm 220 (J/K/mol)
HC(O)Cl(aq) :Delta Hf -376.56 (kJ/mol) Sm 246.52 (J/K/mol)


I started to check up values for HCOOH, SO2 and HCl

HCOOH (aq): Delta Hf -424,72 (kJ/mol) Sm 128,95 (J/K/mol)
HCl(q) :Delta Hf -92,31 (kJ/mol) Sm 186,91 (J/K/mol)
SO2(q) :Delta Hf -296,83 (kJ/mol) Sm 248,22
 (J/K/mol)
HCl(aq) :Delta Hf -167,16 (kJ/mol) Sm 56,5 (J/K/mol)
SO2(aq) : No values

Delta Hf= -376,56-92,31-296,83-(-424,72-785)= 444,02 KJ/mol

Delta S= 246,52+186,91+248,22-128,95-220=332,7 J/K/mol

When delta H and delta S are positive the reaction is entropy driven.

BUT how can I check up if the the driving force is because HCl and SO2 are gases? I cannot find any values for SO2 when it is in liquid phase. 

Offline spirochete

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Re: Carbonyl Compounds and some Thermodynamics
« Reply #7 on: November 07, 2013, 09:15:02 PM »
[quote author=AmbitiousStudent link=topic=71597.msg259123#msg259123 date=1383691916
BUT how can I check up if the the driving force is because HCl and SO2 are gases? I cannot find any values for SO2 when it is in liquid phase.
[/quote]

You don't need to know the exact values. If the product of any reaction is a gas, then this by definition contributes as a driving force and pushes the reaction forward from a thermodynamic perspective. Assuming you're doing the reaction in an open flask, some or all of the gaseous product will bubble out of solution into the surrounding air. This pulls the reaction forward by Le Chatlier's principle.

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