why is the 1-iodobutane omitted in the SN2 reaction with 15% NaI in acetone?
Q. why is the 1-iodobutane omitted in the SN2 reaction with 15% NaI in acetone?
Asked by Dafney - Sat Oct 31 16:36:57 2009 - - 0 Answers - 0 Comments
Q. why is the 1-iodobutane omitted in the SN2 reaction with 15% NaI in acetone?
Asked by Dafney - Sat Oct 31 16:36:57 2009 - - 0 Answers - 0 Comments
For my organic chemistry class, this is the question: what is the major product of this Sn2 reaction?
Q. The reaction is (R)-3-iodobutane + NaI (in an acetone solution) and the want to know the major products. I since the nucleophile is Iodine ion from the Na i believe it replaces the iodine ion on the alkane and changes the sterochemistry to S...but then there is still an Iodine ion floating which once again attacks the alkane and changes it back to R, making it a racemic solution. Am i right or wrong..so confused!?
Asked by Nellab Y - Thu Feb 12 22:28:28 2009 - - 1 Answers - 0 Comments
A. You are correct. This would also be true if you started with the chloride. At first, it would displace to give the S, but slowly, it would racemize exactly as you suggested.
Answered by Dr OChem - Fri Feb 13 00:15:16 2009
Q. The reaction is (R)-3-iodobutane + NaI (in an acetone solution) and the want to know the major products. I since the nucleophile is Iodine ion from the Na i believe it replaces the iodine ion on the alkane and changes the sterochemistry to S...but then there is still an Iodine ion floating which once again attacks the alkane and changes it back to R, making it a racemic solution. Am i right or wrong..so confused!?
Asked by Nellab Y - Thu Feb 12 22:28:28 2009 - - 1 Answers - 0 Comments
A. You are correct. This would also be true if you started with the chloride. At first, it would displace to give the S, but slowly, it would racemize exactly as you suggested.
Answered by Dr OChem - Fri Feb 13 00:15:16 2009
I need help with drawing 3D molecules such as CH3CH2CO2H. Can you please descibe how to draw it.?
Q. I also need help with completing the newman projection and naming them according to Cahn Ingold Prelog rules. The stereoisomers I am using is 2-bromo-3-iodobutane. I don't understand this at all! =( any help will be greatly appreciated
Asked by azn_kazza77 - Fri Apr 18 22:56:16 2008 - - 2 Answers - 0 Comments
A. You have in your hands (literally) one of the simplest tools for visualizing molecules in 3-D. This is one of the first things I teach beginners: make a V or 'peace' sign with your index and middle fingers, then stick out your thumb as well. These three digits represent three of the four apex points of a tetrahedron. The fourth apex is your wrist. Now where your fingers and thumb come together would be the central carbon atom of the 3-D centre you are thinking about. For your example, CH3CH2COOH, that would be the -CH2- carbon atom. If you think of or even mark each of those four points one your hand as being -CH3, -H, -H, and -COOH, and hold your hand as described above you will immediately see the 3-D structure of the molecule. To take… [cont.]
Answered by rmjrenneboog - Sat Apr 19 11:48:10 2008
Q. I also need help with completing the newman projection and naming them according to Cahn Ingold Prelog rules. The stereoisomers I am using is 2-bromo-3-iodobutane. I don't understand this at all! =( any help will be greatly appreciated
Asked by azn_kazza77 - Fri Apr 18 22:56:16 2008 - - 2 Answers - 0 Comments
A. You have in your hands (literally) one of the simplest tools for visualizing molecules in 3-D. This is one of the first things I teach beginners: make a V or 'peace' sign with your index and middle fingers, then stick out your thumb as well. These three digits represent three of the four apex points of a tetrahedron. The fourth apex is your wrist. Now where your fingers and thumb come together would be the central carbon atom of the 3-D centre you are thinking about. For your example, CH3CH2COOH, that would be the -CH2- carbon atom. If you think of or even mark each of those four points one your hand as being -CH3, -H, -H, and -COOH, and hold your hand as described above you will immediately see the 3-D structure of the molecule. To take… [cont.]
Answered by rmjrenneboog - Sat Apr 19 11:48:10 2008
whats the link between bond energy's and reaction rate?
Q. im doing and experiment on: iodobutane bromobutane and chlorobutane i need to know the link between the bond energy and reaction rate how the bond energy has an effect on the reaction rate does it make the reaction faster than others.
Asked by ali300 - Sun Apr 26 21:55:56 2009 - - 2 Answers - 0 Comments
Q. im doing and experiment on: iodobutane bromobutane and chlorobutane i need to know the link between the bond energy and reaction rate how the bond energy has an effect on the reaction rate does it make the reaction faster than others.
Asked by ali300 - Sun Apr 26 21:55:56 2009 - - 2 Answers - 0 Comments
Combustion of halogenoalkane?
Q. How readily do the following burn? 1-chlorobutane 1-bromobutane 1-iodobutane Why is this? cheers
Asked by HM - Fri May 11 12:26:06 2007 - - 1 Answers - 0 Comments
A. In general, halo-alkanes decompose to release halogen atoms that combine readily with active hydrogen atoms, quenching the flame propagation reaction even when adequate fuel, oxygen and heat remains. The chemical reaction in a flame proceeds as a free radical chain reaction; by sequestering the radicals which propagate the reaction, halo-alkanes are able to "poison" the fire at much lower concentrations than are required by fire suppressants using the more traditional methods of cooling, oxygen deprivation, or fuel dilution. ref:
Answered by Dr Dave P - Fri May 11 12:56:19 2007
Q. How readily do the following burn? 1-chlorobutane 1-bromobutane 1-iodobutane Why is this? cheers
Asked by HM - Fri May 11 12:26:06 2007 - - 1 Answers - 0 Comments
A. In general, halo-alkanes decompose to release halogen atoms that combine readily with active hydrogen atoms, quenching the flame propagation reaction even when adequate fuel, oxygen and heat remains. The chemical reaction in a flame proceeds as a free radical chain reaction; by sequestering the radicals which propagate the reaction, halo-alkanes are able to "poison" the fire at much lower concentrations than are required by fire suppressants using the more traditional methods of cooling, oxygen deprivation, or fuel dilution. ref:
Answered by Dr Dave P - Fri May 11 12:56:19 2007
Sn2 and Sn1 rxns?
Q. What is the major organic product for each of the Sn1 & Sn2 rxns: 1. (R)-2-iodobutane reacted w/ sodium propynylide & THF (Sn2 rxn) 2. 1-(2-chloropropyl)-cyclop entane reacted w/ CH3CH2SH (Sn1 rxn) 3. 5-chloro-1-pentanol reacted w/ sodium hydroxide (Sn2 rxn) If you could, explain your answers (be specific) and answer by 4/23/08.
Asked by owlprowler77 - Mon Apr 21 14:41:45 2008 - - 2 Answers - 0 Comments
A. 1. Sn2 reactions procedd with inversion of stereochemistry about the reactive centre so you will need to assess the order of constituents in the product to assign it as R- or S-. The propynylide ion will replace the iodide ion in the product to give CH3CH(CCCH3)CH2CH3 (4-methyl-2-hexyne) 2 Sn1 reactions proceed through a symmetrical carbonium ion intermediate, so you should expect equal quantities of R- and S- 1-(2-thioethyl)-cyclopent ane as the -SCH2CH3 group replaces the Cl- substituent. 3. Sn2 reaction should give you simply 1,5-pentanediol. No stereochemistry applies since the reaction is not producing an asymmetrical centre.
Answered by rmjrenneboog - Mon Apr 21 15:00:04 2008
Q. What is the major organic product for each of the Sn1 & Sn2 rxns: 1. (R)-2-iodobutane reacted w/ sodium propynylide & THF (Sn2 rxn) 2. 1-(2-chloropropyl)-cyclop entane reacted w/ CH3CH2SH (Sn1 rxn) 3. 5-chloro-1-pentanol reacted w/ sodium hydroxide (Sn2 rxn) If you could, explain your answers (be specific) and answer by 4/23/08.
Asked by owlprowler77 - Mon Apr 21 14:41:45 2008 - - 2 Answers - 0 Comments
A. 1. Sn2 reactions procedd with inversion of stereochemistry about the reactive centre so you will need to assess the order of constituents in the product to assign it as R- or S-. The propynylide ion will replace the iodide ion in the product to give CH3CH(CCCH3)CH2CH3 (4-methyl-2-hexyne) 2 Sn1 reactions proceed through a symmetrical carbonium ion intermediate, so you should expect equal quantities of R- and S- 1-(2-thioethyl)-cyclopent ane as the -SCH2CH3 group replaces the Cl- substituent. 3. Sn2 reaction should give you simply 1,5-pentanediol. No stereochemistry applies since the reaction is not producing an asymmetrical centre.
Answered by rmjrenneboog - Mon Apr 21 15:00:04 2008
Organic Chemistry please help!!?
Q. Iodide ion can replace the iodine of an alkyl iodide by a substitution reaction. For example, the reaction of 131I- (radioactive iodide ion, I*-) with 2-iodobutane can be monitored by measuring the increasing amount of radioactivity incorporated into the 2-iodobutane: CH3CHICH2CH3 + I*- CH3CHI*CH2CH3 + I- If the experiment is carried out with a pure enantiomer of the alkyl iodide, for example (R)-2-iodobutane, the rate of the reaction can also be monitored by continuously measuring the specific rotation, and watching how rapidly racemization takes place (how rapidly [ ]D goes to zero). It turns out that the initial rate of incorporation of radioactivity (on a mole per mole basis) is exactly half the rate at which the specific rotation… [cont.]
Asked by Henry S - Mon Oct 13 17:56:23 2008 - - 2 Answers - 0 Comments
A. Think it through. If the reaction is SN2, each substitution will be accompanied by an inversion. However, once the reaction has been going for some time, some material that has already reacted will react again, complicating matters. If the reaction is Sn1, you will have a planar intermediate, so reaction will lead to inversion or retention with equal probability.
Answered by Paul B - Mon Oct 13 18:06:13 2008
Q. Iodide ion can replace the iodine of an alkyl iodide by a substitution reaction. For example, the reaction of 131I- (radioactive iodide ion, I*-) with 2-iodobutane can be monitored by measuring the increasing amount of radioactivity incorporated into the 2-iodobutane: CH3CHICH2CH3 + I*- CH3CHI*CH2CH3 + I- If the experiment is carried out with a pure enantiomer of the alkyl iodide, for example (R)-2-iodobutane, the rate of the reaction can also be monitored by continuously measuring the specific rotation, and watching how rapidly racemization takes place (how rapidly [ ]D goes to zero). It turns out that the initial rate of incorporation of radioactivity (on a mole per mole basis) is exactly half the rate at which the specific rotation… [cont.]
Asked by Henry S - Mon Oct 13 17:56:23 2008 - - 2 Answers - 0 Comments
A. Think it through. If the reaction is SN2, each substitution will be accompanied by an inversion. However, once the reaction has been going for some time, some material that has already reacted will react again, complicating matters. If the reaction is Sn1, you will have a planar intermediate, so reaction will lead to inversion or retention with equal probability.
Answered by Paul B - Mon Oct 13 18:06:13 2008
Organic Reactions?
Q. Please help me ASAP as to what are the products and Formulas of each molecule of each rxn. Substitution Reactions 1. 1-iodobutane + Potassium hydroxide ---> 2. Tetrachloromethane + fluorine ---> Addition Reactions 1. Ethene +chlorine ---> 2.Ethene + water ---> 3.Propene + hydrogen ---> 4. 1,4 hexadiene +iodine(in excess) ---> Elimination Reactions 1.Ethanol --->(h2so4 over arrow in all elimination rxns) 2. 1-aminobutane---> 3. 2,3-dichloropentane---> Condensation rxns 1. butanol +ethanoic acid ---> 2. ethanol + pentanoic acid ---> Addition Polymerization 1. propene ---> 2. chloroethene ---> 3. tetrafluoroethene ---> Condensation Polymerization 1.4-hydroxypentanoic acid ---> 2.6-aminohexanoic acid ---> 3. 1,2-ethandiol + butandioic acid --- [cont.]
Asked by Michael F - Tue May 20 20:37:56 2008 - - 1 Answers - 0 Comments
Q. Please help me ASAP as to what are the products and Formulas of each molecule of each rxn. Substitution Reactions 1. 1-iodobutane + Potassium hydroxide ---> 2. Tetrachloromethane + fluorine ---> Addition Reactions 1. Ethene +chlorine ---> 2.Ethene + water ---> 3.Propene + hydrogen ---> 4. 1,4 hexadiene +iodine(in excess) ---> Elimination Reactions 1.Ethanol --->(h2so4 over arrow in all elimination rxns) 2. 1-aminobutane---> 3. 2,3-dichloropentane---> Condensation rxns 1. butanol +ethanoic acid ---> 2. ethanol + pentanoic acid ---> Addition Polymerization 1. propene ---> 2. chloroethene ---> 3. tetrafluoroethene ---> Condensation Polymerization 1.4-hydroxypentanoic acid ---> 2.6-aminohexanoic acid ---> 3. 1,2-ethandiol + butandioic acid --- [cont.]
Asked by Michael F - Tue May 20 20:37:56 2008 - - 1 Answers - 0 Comments
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