Test 3 OCHEM
Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with HBr
(1-bromoethyl)cyclohexane
Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with HCl
(1-chloroethyl)cyclohexane
Use ChemSketch to draw the alkene constitutional and stereoisomers of C4H7Cl
(1E)-1-chlorobut-1-ene (1Z)-1-chlorobut-1-ene 2-chlorobut-1-ene 3-chlorobut-1-ene 4-chlorobut-1-ene (2Z)-1-chlorobut-2-ene (2E)-1-chlorobut-2-ene (2Z)-2-chlorobut-2-ene (2E)-2-chlorobut-2-ene 1-chloro-2-methylprop-1-ene 3-chloro-2-methylprop-1-ene
Cortisone (C21H28O5) is an injectable steroid used to treat inflammation in tendons, bursa, and joints. Calculate the degrees unsaturation.
8
15. Use the first stereochemical style shown in the picture below to draw the two stereoisomers (i.e. where the first group that adds to the double bond can add from the front or back side of the molecule and the other group will add according to the mechanism (syn or anti) - producing two compounds and two answers) that result from the following reaction: 1-methylcyclohexene SMILES = CC=1CCCCC=1 + Br2 → ________ + ________
(1R,2R)-1,2-dibromo-1-methylcyclohexane (1S,2S)-1,2-dibromo-1-methylcyclohexane
16. Use the first stereochemical style shown in the picture below to draw the two stereoisomers (i.e. where the first group that adds to the double bond can add from the front or back side of the molecule and the other group will add according to the mechanism (syn or anti) - producing two compounds and two answers) that result from the following reaction: 1-methylcyclohexene SMILES = CC=1CCCCC=1 + Br2 / H2O → ________ + ________
(1R,2R)-2-bromo-1-methylcyclohexanol (1S,2S)-2-bromo-1-methylcyclohexanol
Use the first stereochemical style shown in the picture below to draw the two stereoisomers (i.e. where the first group that adds to the double bond can add from the front or back side of the molecule and the other group will add according to the mechanism (syn or anti) - producing two compounds and two answers) that result from the following reaction: 1-methylcyclohexene SMILES = CC=1CCCCC=1 + BD3 then H2O2 / OH- → ________ + ________
(1R,2R)-2-methyl(2-2H)cyclohexanol (1S,2S)-2-methyl(2-2H)cyclohexanol
Use the first stereochemical style shown in the picture below to draw the two stereoisomers (i.e. where the first group that adds to the double bond can add from the front or back side of the molecule and the other group will add according to the mechanism (syn or anti) - producing two compounds and two answers) that result from the following reaction: 1-methylcyclohexene SMILES = CC=1CCCCC=1 + BH3 then H2O2 / OH- → ________ + ________
(1R,2R)-2-methylcyclohexanol (1S,2S)-2-methylcyclohexanol
Use the first stereochemical style shown in the picture below to draw the two stereoisomers (i.e. where the first group that adds to the double bond can add from the front or back side of the molecule and the other group will add according to the mechanism (syn or anti) - producing two compounds and two answers) that result from the following reaction: 1-methylcyclohexene SMILES = CC=1CCCCC=1 + D2 / Pd(C) or PtO2 → ________ + ________
(1R,2R)-methyl(1,2-2H2)cyclohexane (1S,2S)-methyl(1,2-2H2)cyclohexane
Use the first stereochemical style shown in the picture below to draw the two stereoisomers (i.e. where the first group that adds to the double bond can add from the front or back side of the molecule and the other group will add according to the mechanism (syn or anti) - producing two compounds and two answers) that result from the following reaction: 1-methylcyclohexene SMILES = CC=1CCCCC=1 + CH2I2 /Zn(Cu) → ________ + ________
(1R,6S)-1-methylbicyclo[4.1.0]heptane (1S,6R)-1-methylbicyclo[4.1.0]heptane
Use the first stereochemical style shown in the picture below to draw the two stereoisomers (i.e. where the first group that adds to the double bond can add from the front or back side of the molecule and the other group will add according to the mechanism (syn or anti) - producing two compounds and two answers) that result from the following reaction: 1-methylcyclohexene SMILES = CC=1CCCCC=1 + OsO4 then NaHSO3 / H2O → ________ + ________
(1S,2R)-1-methylcyclohexane-1,2-diol (1R,2S)-1-methylcyclohexane-1,2-diol
The first step in the reaction below adds Hydrogen to one of the double bond carbons. This forms a trigonal planar carbocation at the adjacent double bond carbon. In the second (and final) step of the mechanism, the nucleophile adds to either the top or bottom lobe of the carbocation's unhybridized p orbital. Overall addition across the double bond is 50% syn and 50% anti - in other words no stereochemical preference in this reaction.Use the first stereochemical style shown in the picture below to draw the two geometric stereoisomers (i.e. where the two methyl groups have a cis / trans relationship in the two answers) that result from the following reaction: 1,2-dimethylcyclohexene SMILES = CC=1CCCCC=1C + H2O / H+ → ________ + ________
(1S,2S)-1,2-dimethylcyclohexanol (1R,2S)-1,2-dimethylcyclohexanol
The first step in the reaction below adds Mercury across the double bond to form a mercurium ion - a 3-membered ring containing a positively charged Mercury atom. Then, water adds to the more substituted carbon in the 3-membered ring and breaks the ring between that carbon and the Hg atom. At this point, anti addition of Hg and OH have occurred. In the final step, NaBH4 is added to replace the Hg atom with a Hydrogen atom - this reaction occurs in such a way that H and OH are on the same side of the molecule 50% of the time (i.e. syn addition) and on opposite sides of the molecule 50% of the time (i.e. anti addition). So, no stereochemical preference is observed in this reaction.Use the first stereochemical style shown in the picture below to draw the two geometric stereoisomers (i.e. where the two methyl groups have a cis / trans relationship in the two answers) that result from the following reaction: 1,2-dimethylcyclohexene SMILES = CC=1CCCCC=1C + Hg(O2CCH3)2 / H2O then NaBH4 → ________ + ________
(1S,2S)-1,2-dimethylcyclohexanol (1R,2S)-1,2-dimethylcyclohexanol
The first step in the reaction below adds Hydrogen to one of the double bond carbons. This forms a trigonal planar carbocation at the adjacent double bond carbon. In the second (and final) step of the mechanism, the nucleophile adds to either the top or bottom lobe of the carbocation's unhybridized, trigonal planar, p orbital. Overall addition across the double bond is 50% syn and 50% anti - in other words no stereochemical preference in this reaction.Use the first stereochemical style shown in the picture below to draw the two geometric stereoisomers (i.e. where the two methyl groups have a cis / trans relationship in the two answers) that result from the following reaction: 1,2-dimethylcyclohexene SMILES = CC=1CCCCC=1C + HCl → ________ + ________
(1S,2S)-1-chloro-1,2-dimethylcyclohexane (1R,2S)-1-chloro-1,2-dimethylcyclohexane
The first step in the reaction below adds Hydrogen to one of the double bond carbons. This forms a trigonal planar carbocation at the adjacent double bond carbon. In the second (and final) step of the mechanism, the nucleophile adds to either the top or bottom lobe of the carbocation's unhybridized p orbital. Overall addition across the double bond is 50% syn and 50% anti - in other words no stereochemical preference in this reaction.Use one of the stereochemical styles shown in the picture below to draw the two geometric stereoisomers (i.e. where the two methyl groups have a cis / trans relationship in the two answers) that result from the following reaction: 1,2-dimethylcyclohexene SMILES = CC=1CCCCC=1C + KI / H3PO4 → ________ + ________
(1S,2S)-1-iodo-1,2-dimethylcyclohexane (1R,2S)-1-iodo-1,2-dimethylcyclohexane
17. Use the first stereochemical style shown in the picture below to draw the two stereoisomers (i.e. where the first group that adds to the double bond can add from the front or back side of the molecule and the other group will add according to the mechanism (syn or anti) - producing two compounds and two answers) that result from the following reaction: 1-methylcyclohexene SMILES = CC=1CCCCC=1 + CHCl3 / KOH → ________ + ________
(1S,6R)-7,7-dichloro-1-methylbicyclo[4.1.0]heptane (1R,6S)-7,7-dichloro-1-methylbicyclo[4.1.0]heptane
Use ChemSketch to draw the Most Stable Butene isomer (or stereoisomer) of C4H8
(2E)-but-2-ene
Use ChemSketch to draw the Most Stable alkene isomer (or stereoisomer) of C4H8
(2E)-but-2-ene
Use ChemSketch to draw the Most Stable alkene isomer (or stereoisomer) of 2-hexene
(2E)-hex-2-ene
Use ChemSketch to generate the names of the C6H12 alkene isomers that CAN NOT show Markovnikov addition of HCl.
(2Z)-4-methylpent-2-ene (2E)-4-methylpent-2-ene 2,3-dimethylbut-2-ene (2Z)-hex-2-ene (2E)-hex-2-ene (3E)-hex-3-ene (3Z)-hex-3-ene
Use ChemSketch to generate the names of the C4H8 alkene isomers that CAN NOT show Markovnikov addition of HCl.
(2Z)-but-2-ene (2E)-but-2-ene
Use ChemSketch to generate the names of the C5H10 alkene isomers that CAN NOT show Markovnikov addition of HCl.
(2Z)-pent-2-ene (2E)-pent-2-ene
Use ChemSketch to draw the Least Stable alkene isomer (or stereoisomer) of 3-hexene Use the lasso tool to select the structure then enter its name below.
(3Z)-hex-3-ene
The elimination of HOH (water) from adjacent carbons in an Alcohol is accomplished by adding H2SO4. A major problem with this reaction is that mixtures of alkenes may be formed (mixtures are undesireable from time and financial perspectives). Enter the names of the possible alkene products that can be formed when the following alcohol reacts with Sulfuric acid. 4-ethyloctan-4-ol SMILES = CCCC(O)(CCCC)CC + H2SO4 →
(4E)-4-ethylideneoctane (4Z)-4-ethylideneoctane (3E)-4-ethyloct-3-ene (3Z)-4-ethyloct-3-ene (4Z)-4-ethyloct-4-ene (4E)-4-ethyloct-4-ene
What diene would you react with 2 equivalents HBr to exclusively form the alkyl dibromide shown below?Draw the diene in ChemSketch and enter its name below. If the alkyl dibromide cannot be formed (or formed exclusively) from any diene using just 2 equivalents HBr, enter the word "None" - without the quotes.
(4E)-4-tert-butyl-3,3-dimethylhexa-1,4-diene
Applying Cahn-Ingold-Prelog Priority Rules appropriately to -CH2CH3 gives the following set of numbers: 6611111Enter the numbers you should obtain for the following substituent: -CHBrCH3
63561111
How many different alkenes can react with HBr to produce the alkyl bromide shown below as the major product? The "major product" is more than 50% of the product mixture. The alkene reactant(s) is(are) drawn by removing the Br and an H from a Carbon adjacent to the Carbon bonded to the Br. This reaction follows Markovnikov Addition so make sure that when the alkene reactant(s) from Step 2 adds HBr that you will indeed get the product shown. Count cis / trans and E / Z isomers separately. Assume that rearrangement DOES NOT OCCUR. TERM ENGLISH
1
What diene would you react with 2 equivalents HBr to exclusively form the alkyl dibromide shown below?Draw the diene in ChemSketch and enter its name below. If the alkyl dibromide cannot be formed (or formed exclusively) from any diene using just 2 equivalents HBr, enter the word "None" - without the quotes.
1,1,3,3,5,5-hexamethyl-2,4-dimethylidenecyclohexane
What alkene would you react with HBr to exclusively form the alkyl bromide shown below?Draw the alkene in ChemSketch and enter its name below. If the alkyl bromide cannot be formed (or formed exclusively) from any alkene using just HBr, enter the word "None" - without the quotes.
1,1,3-trimethyl-2-methylidenecyclopentane
same as above
1,2-dimethylcyclobutylium
same as above
1-(1-bromo-1-methylethyl)-1-methylcyclopentane
same as above
1-(2-bromopropyl)-1-methylcyclopentane
same as above
1-bromo-1,2-dimethylcyclohexane
same as above
1-bromo-1-(2,2-dimethylpropyl)cyclopentane
Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with HBr
1-bromo-1-ethylcyclohexane
same as above
1-bromo-1-ethylcyclopentane
Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with HBr
1-bromo-1-methylcyclohexane
Deuterium (symbol = D) is an isotope of Hydrogen. In fact, there are 3 isotopes of Hydrogen: Protium, Deuterium and Tritium. Protium (1 proton, 0 neutrons) is the most prevalent isotope of Hydrogen making up 99.98 % of all atoms of Hydrogen. Deuterium (1 proton, 1 neutron) is used to make "heavy water", D2O, which is used to moderate (slow down) neutrons in nuclear reactors. Tritium (1 proton, 2 neutrons) is radioactive and undergoes a fusion reaction with deuterium in the modern thermonuclear weapon (aka Hydrogen Bomb). Chemists add Deuterium to organic compounds using the very same reactions one would use to add Hydrogen - since Deuterium IS Hydrogen. Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with DCL
1-chloro-1-(2H1)methylcyclohexane
Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with HCl
1-chloro-1-ethylcyclohexane
Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with HCl
1-chloro-1-methylcyclohexane
The elimination of HOH (water) from adjacent carbons in an Alcohol is accomplished by adding H2SO4. A major problem with this reaction is that mixtures of alkenes may be formed (mixtures are undesireable from time and financial perspectives). Enter the names of the possible alkene products that can be formed when the following alcohol reacts with Sulfuric acid. 1-ethyl-3-methylcyclopentanol SMILES = CC1CCC(O)(CC)C1 + H2SO4 →
1-ethyl-3-methylcyclopentene 1-ethyl-4-methylcyclopentene (1E)-1-ethylidene-3-methylcyclopentane (1Z)-1-ethylidene-3-methylcyclopentane
The elimination of HOH (water) from adjacent carbons in an Alcohol is accomplished by adding H2SO4. A major problem with this reaction is that mixtures of alkenes may be formed (mixtures are undesireable from time and financial perspectives). Enter the names of the possible alkene products that can be formed when the following alcohol reacts with Sulfuric acid. 1-ethyl-2-methylcyclopentanol SMILES = CC1CCCC1(O)CC + H2SO4 →
1-ethyl-5-methylcyclopentene 1-ethyl-2-methylcyclopentene (1Z)-1-ethylidene-2-methylcyclopentane (1E)-1-ethylidene-2-methylcyclopentane
Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with KI / H3PO4
1-iodo-1-methylcyclohexane
The elimination of HBr (an acid) from adjacent carbons in an Alkyl Bromide is accomplished by adding KOH (a base) dissolved in ethanol. A major problem with this reaction is that mixtures of alkenes may be formed (mixtures are undesireable from time and financial perspectives). Enter the names of the possible alkene products that can be formed when the following alkyl bromide reacts with KOH / ethanol. 1-bromo-1,2-dimethylcyclohexane + KOH →
1-methyl-2-methylidenecyclohexane 1,2-dimethylcyclohexene 1,6-dimethylcyclohexene
same as above
1-methylcyclobutylium
C7H?ClO2N contains a ring, a double bond (i.e. an alkene), an amide and an alcohol (these functional groups represent ALL the ° Unsaturation). How many Hydrogen atoms are present?
10
C8H?ClO2N contains a ring, a nitrile and an ester (these functional groups represent ALL the ° Unsaturation). How many Hydrogen atoms are present?
10
Heroin (C21H23NO5) is an alkaloid present in the opium poppy, Papaver somniferum. Once heroin enters the bloodstream, it quickly crosses the blood-brain barrier where it is converted to morphine. Calculate the degrees unsaturation.
11
How many different alkenes can react with HBr to produce the alkyl bromide shown below as the major product? The "major product" is more than 50% of the product mixture. The alkene reactant(s) is(are) drawn by removing the Br and an H from a Carbon adjacent to the Carbon bonded to the Br. This reaction follows Markovnikov Addition so make sure that when the alkene reactant(s) from Step 2 adds HBr that you will indeed get the product shown. Count cis / trans and E / Z isomers separately. Assume that rearrangement DOES NOT OCCUR.
2
Use ChemSketch to draw the Most Stable alkene isomer (or stereoisomer) of C6H12
2,3-dimethylbut-2-ene
Carbocations rearrange to more stable carbocations by one of the following mechanisms: Hydride Shift - a hydrogen with its pair of bonding electrons (essentially a hydride ion) can move to an adjacent carbon atom if a more stable carbocation results. The same carbon skeleton remains after a Hydride Shift, but the positive charge is located on a different carbon atom than expected. Alkyl Shift - an alkyl group (even an alkyl portion of a ring) with its pair of bonding electrons can move to an adjacent carbon atom if a more stable carbocation results. A different carbon skeleton results after an Alkyl Shift, and both the alkyl group and the positive charge are located on a different carbon atoms than expected.If an Alkyl Shift occurs in a ring, a larger ring will form ("ring-expansion"). If the Alkyl Shift can be a normal alkyl group or a ring-expansion, show the ring-expansion product. If both the Hydride Shift and the Alkyl Shift are possible, show the one that forms the MOST stable carbocation. Use ChemSketch to draw the rearranged carbocation of the structure below. If a rearrangement is not likely to occur (i.e. the molecule is not structured to take advantage of either of the two shifts mentioned above), enter "None" without the quotes.
2,3-dimethylbut-2-ylium
Use ChemSketch to draw the Most Stable alkene isomer (or stereoisomer) of C7H14
2,3-dimethylpent-2-ene
Deuterium (symbol = D) is an isotope of Hydrogen. In fact, there are 3 isotopes of Hydrogen: Protium, Deuterium and Tritium. Protium (1 proton, 0 neutrons) is the most prevalent isotope of Hydrogen making up 99.98 % of all atoms of Hydrogen. Deuterium (1 proton, 1 neutron) is used to make "heavy water", D2O, which is used to moderate (slow down) neutrons in nuclear reactors. Tritium (1 proton, 2 neutrons) is radioactive and undergoes a fusion reaction with deuterium in the modern thermonuclear weapon (aka Hydrogen Bomb). Chemists add Deuterium to organic compounds using the very same reactions one would use to add Hydrogen - since Deuterium IS Hydrogen. Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with 2 Equivalents of DCl
2,4-dichloro(1,5-2H2)pentane
same as above
2-bromo-1,1-dimethylcyclohexane
The addition of HBr to an alkene proceeds through a carbocation intermediate. Once formed, carbocations can rearrange to more stable carbocations by one of the following mechanisms: Hydride Shift - a hydrogen with its pair of bonding electrons (essentially a hydride ion) can move to an adjacent carbon atom if a more stable carbocation results. The same carbon skeleton remains after a Hydride Shift, but the positive charge is located on a different carbon atom than expected. Alkyl Shift - an alkyl group (even an alkyl portion of a ring) with its pair of bonding electrons can move to an adjacent carbon atom if a more stable carbocation results. A different carbon skeleton results after an Alkyl Shift, and both the alkyl group and the positive charge are located on a different carbon atoms than expected.If an Alkyl Shift occurs in a ring, a larger ring will form ("ring-expansion"). If the Alkyl Shift can be a normal alkyl group or a ring-expansion, show the ring-expansion product. If both the Hydride Shift and the Alkyl Shift are possible, show the one that forms the MOST stable carbocation. Use ChemSketch to draw the alkyl bromide product formed when HBr reacts with the alkene shown - enter the ChemSketch-generated name below. A rearrangement may not occur in all cases (i.e. the molecule is not structured to take advantage of either of the two shifts mentioned above), but the addition of HBr will occur.
2-bromo-2,3-dimethylbutane
same as above
2-bromo-2,4,4-trimethylpentane
same as above
2-bromo-2-methylbutane
Deuterium (symbol = D) is an isotope of Hydrogen. In fact, there are 3 isotopes of Hydrogen: Protium, Deuterium and Tritium. Protium (1 proton, 0 neutrons) is the most prevalent isotope of Hydrogen making up 99.98 % of all atoms of Hydrogen. Deuterium (1 proton, 1 neutron) is used to make "heavy water", D2O, which is used to moderate (slow down) neutrons in nuclear reactors. Tritium (1 proton, 2 neutrons) is radioactive and undergoes a fusion reaction with deuterium in the modern thermonuclear weapon (aka Hydrogen Bomb). Chemists add Deuterium to organic compounds using the very same reactions one would use to add Hydrogen - since Deuterium IS Hydrogen. Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with DCl
2-chloro(1-2H1)pentane
Deuterium (symbol = D) is an isotope of Hydrogen. In fact, there are 3 isotopes of Hydrogen: Protium, Deuterium and Tritium. Protium (1 proton, 0 neutrons) is the most prevalent isotope of Hydrogen making up 99.98 % of all atoms of Hydrogen. Deuterium (1 proton, 1 neutron) is used to make "heavy water", D2O, which is used to moderate (slow down) neutrons in nuclear reactors. Tritium (1 proton, 2 neutrons) is radioactive and undergoes a fusion reaction with deuterium in the modern thermonuclear weapon (aka Hydrogen Bomb). Chemists add Deuterium to organic compounds using the very same reactions one would use to add Hydrogen - since Deuterium IS Hydrogen. Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with DCl
2-chloro-2-methyl(1-2H1)propane
Deuterium (symbol = D) is an isotope of Hydrogen. In fact, there are 3 isotopes of Hydrogen: Protium, Deuterium and Tritium. Protium (1 proton, 0 neutrons) is the most prevalent isotope of Hydrogen making up 99.98 % of all atoms of Hydrogen. Deuterium (1 proton, 1 neutron) is used to make "heavy water", D2O, which is used to moderate (slow down) neutrons in nuclear reactors. Tritium (1 proton, 2 neutrons) is radioactive and undergoes a fusion reaction with deuterium in the modern thermonuclear weapon (aka Hydrogen Bomb). Chemists add Deuterium to organic compounds using the very same reactions one would use to add Hydrogen - since Deuterium IS Hydrogen. Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with DCl
2-chloro-2-methyl(3-2H1)butane
( ch 7A, question 1) The elimination of HBr (an acid) from adjacent carbons in an Alkyl Bromide is accomplished by adding KOH (a base) dissolved in ethanol. A major problem with this reaction is that mixtures of alkenes may be formed (mixtures are undesireable from time and financial perspectives). Enter the names of the possible alkene products that can be formed when the following alkyl bromide reacts with KOH / ethanol. 2-bromo-2-methylbutane + KOH →
2-methylbut-1-ene 2-methylbut-2-ene
Carbocations rearrange to more stable carbocations by one of the following mechanisms: Hydride Shift - a hydrogen with its pair of bonding electrons (essentially a hydride ion) can move to an adjacent carbon atom if a more stable carbocation results. The same carbon skeleton remains after a Hydride Shift, but the positive charge is located on a different carbon atom than expected. Alkyl Shift - an alkyl group (even an alkyl portion of a ring) with its pair of bonding electrons can move to an adjacent carbon atom if a more stable carbocation results. A different carbon skeleton results after an Alkyl Shift, and both the alkyl group and the positive charge are located on a different carbon atoms than expected.If an Alkyl Shift occurs in a ring, a larger ring will form ("ring-expansion"). If the Alkyl Shift can be a normal alkyl group or a ring-expansion, show the ring-expansion product. If both the Hydride Shift and the Alkyl Shift are possible, show the one that forms the MOST stable carbocation. Use ChemSketch to draw the rearranged carbocation of the structure below. If a rearrangement is not likely to occur (i.e. the molecule is not structured to take advantage of either of the two shifts mentioned above), enter "None" without the quotes.
2-methylbut-2-ylium
Use ChemSketch to draw the Most Stable alkene isomer (or stereoisomer) of C5H10
2-methylbut-e-ene
same as above
3-bromo-1-ethyl-1-methylcyclopentane
What alkene would you react with HBr to exclusively form the alkyl bromide shown below?Draw the alkene in ChemSketch and enter its name below. If the alkyl bromide cannot be formed (or formed exclusively) from any alkene using just HBr, enter the word "None" - without the quotes.
3-ethyl-2-methylhex-1-ene
What alkene would you react with HBr to exclusively form the alkyl bromide shown below?Draw the alkene in ChemSketch and enter its name below. If the alkyl bromide cannot be formed (or formed exclusively) from any alkene using just HBr, enter the word "None" - without the quotes.
3-ethyl-4-methylpent-2-ene
Use ChemSketch to generate the names of the C6H10 Cyclobutene isomers that CAN NOT show Markovnikov addition of HCl.
3-ethylcyclobutene 3,3-dimethylcyclobutene 3,4-dimethylcyclobutene 1,2-dimethylcyclobutene
The elimination of HBr (an acid) from adjacent carbons in an Alkyl Bromide is accomplished by adding KOH (a base) dissolved in ethanol. A major problem with this reaction is that mixtures of alkenes may be formed (mixtures are undesireable from time and financial perspectives). Enter the names of the possible alkene products that can be formed when the following alkyl bromide reacts with KOH / ethanol. 3-bromo-3-methylhexane + KOH →
3-methylidenehexane (2E)-3-methylhex-2-ene (2Z)-3-methylhex-2-ene (3E)-3-methylhex-3-ene (3Z)-3-methylhex-3-ene
The elimination of HBr (an acid) from adjacent carbons in an Alkyl Bromide is accomplished by adding KOH (a base) dissolved in ethanol. A major problem with this reaction is that mixtures of alkenes may be formed (mixtures are undesireable from time and financial perspectives). Enter the names of the possible alkene products that can be formed when the following alkyl bromide reacts with KOH / ethanol. 2-bromo-3-methylpentane + KOH →
3-methylpent-1-ene (2E)-3-methylpent-2-ene (2Z)-3-methylpent-2-ene
Use ChemSketch to generate the names of the C6H10 Cyclopentene isomers that CAN NOT show Markovnikov addition of HCl.
4-methylcyclopentene 3-methylcyclopentene
Arrange the following groups according to their priority using the Cahn-Ingold-Prelog sequence rules
Br Cl SH OCH2CH3 NHCH3 NH2 CONH2 CH=CH2 CH2CH3 CH3
Use ChemSketch to draw (2Z,4E)-3-tert-butyl-4-ethyl-2,4-hexadiene Use the lasso tool to select the structure then Click Tools / Generate / SMILES Notation to produce a SMILE (Simplified Molecular Input Line Entry) for the molecule. Paste the SMILE below.
CC/C(=C\C)/C(=C\C)C(C)(C)C
Use ChemSketch to draw (2E,4Z,6Z)-3-ethyl-5-isopropyl-6-propyl-2,4,6-nonatriene Use the lasso tool to select the structure then Click Tools / Generate / SMILES Notation to produce a SMILE (Simplified Molecular Input Line Entry) for the molecule. Paste the SMILE below.
CCC\C(=C\CC)\C(=C/C(=C/C)CC)C(C)C
Use ChemSketch to draw (2Z,4E,6Z)-3-ethyl-5-isopropyl-6-propyl-2,4,6-nonatriene Use the lasso tool to select the structure then Click Tools / Generate / SMILES Notation to produce a SMILE (Simplified Molecular Input Line Entry) for the molecule. Paste the SMILE below.
CCC\C(=C\CC)\C(=C\C(=C/C)CC)C(C)C
Use ChemSketch to draw (3Z,5Z)-3-tert-butyl-5-ethyl-1,3,5-octatriene Use the lasso tool to select the structure then Click Tools / Generate / SMILES Notation to produce a SMILE (Simplified Molecular Input Line Entry) for the molecule. Paste the SMILE below.
CC\C(\C=C(\C=C)C(C)(C)C)=C\CC
Use ChemSketch to draw (2Z,4Z,6Z)-5-tert-butyl-3-ethyl-2,4,6-nonatriene Use the lasso tool to select the structure then Click Tools / Generate / SMILES Notation to produce a SMILE (Simplified Molecular Input Line Entry) for the molecule. Paste the SMILE below.
CC\C(\C=C(\C=C/CC)C(C)(C)C)=C\C
Use ChemSketch to draw (2Z,4E,6Z)-6-ethyl-4-methyl-2,4,6-nonatriene Use the lasso tool to select the structure then Click Tools / Generate / SMILES Notation to produce a SMILE (Simplified Molecular Input Line Entry) for the molecule. Paste the SMILE below.
C\C(=C/C(=C\CC)CC)\C=C/C
Use ChemSketch to draw (4E,7Z)-2-methyl-2,4,7-nonatriene Use the lasso tool to select the structure then Click Tools / Generate / SMILES Notation to produce a SMILE (Simplified Molecular Input Line Entry) for the molecule. Paste the SMILE below.
C\C=C/C\C=C\C=C(\C)C
Use ChemSketch to draw (4E,7E)-2,4,6-trimethyl-2,4,7-nonatriene Use the lasso tool to select the structure then Click Tools / Generate / SMILES Notation to produce a SMILE (Simplified Molecular Input Line Entry) for the molecule. Paste the SMILE below.
C\C=C\C(C)/C=C(\C)/C=C(\C)C
Use Cahn-Prelog-Ingold's sequence rules to determine whether the structure below is E or Z or Has no stereoisomer.
Z
Use ChemSketch to draw the Least Stable alkene isomer (or stereoisomer) of C4H8
but-1-ene
The elimination of HBr (an acid) from adjacent carbons in an Alkyl Bromide is accomplished by adding KOH (a base) dissolved in ethanol. A major problem with this reaction is that mixtures of alkenes may be formed (mixtures are undesireable from time and financial perspectives). Enter the names of the possible alkene products that can be formed when the following alkyl bromide reacts with KOH / ethanol. 2-bromobutane + KOH →
but-1-ene (2E)-but-2-ene (2Z)-but-2-ene
Carbocations rearrange to more stable carbocations by one of the following mechanisms: Hydride Shift - a hydrogen with its pair of bonding electrons (essentially a hydride ion) can move to an adjacent carbon atom if a more stable carbocation results. The same carbon skeleton remains after a Hydride Shift, but the positive charge is located on a different carbon atom than expected. Alkyl Shift - an alkyl group (even an alkyl portion of a ring) with its pair of bonding electrons can move to an adjacent carbon atom if a more stable carbocation results. A different carbon skeleton results after an Alkyl Shift, and both the alkyl group and the positive charge are located on a different carbon atoms than expected.If an Alkyl Shift occurs in a ring, a larger ring will form ("ring-expansion"). If the Alkyl Shift can be a normal alkyl group or a ring-expansion, show the ring-expansion product. If both the Hydride Shift and the Alkyl Shift are possible, show the one that forms the MOST stable carbocation. Use ChemSketch to draw the rearranged carbocation of the structure below. If a rearrangement is not likely to occur (i.e. the molecule is not structured to take advantage of either of the two shifts mentioned above), enter "None" without the quotes.
but-2-ylium
Use ChemSketch to draw the major organic product (review Markovnikov Addition) when the compound pictured below reacts with KI / H3PO4
equal
Use Cahn-Prelog-Ingold's sequence rules to determine whether the structure below is E or Z or Has no stereoisomer.
has no stereoisomer
Use ChemSketch to draw the Least Stable Heptene isomer (or stereoisomer) of C7H14
hept-1-ene
Use ChemSketch to draw the Least Stable Hexene isomer (or stereoisomer) of C6H12
hex-1-ene
What alkene would you react with HBr to exclusively form the alkyl bromide shown below?Draw the alkene in ChemSketch and enter its name below. If the alkyl bromide cannot be formed (or formed exclusively) from any alkene using just HBr, enter the word "None" - without the quotes.
hex-1-ene
Use ChemSketch to generate the names of the C6H12 alkene isomers that CAN NOT have a cis / trans (or E / Z) stereoisomer.
hex-1-ene 3-methylidenepentane 2-methylpent-1-ene 2,3-dimethylbut-1-ene 2-methylpent-2-ene 2,3-dimethylbut-2-ene 3,3-dimethylbut-1-ene 3-methylpent-1-ene 4-methylpent-1-ene
Carbocations rearrange to more stable carbocations by one of the following mechanisms: Hydride Shift - a hydrogen with its pair of bonding electrons (essentially a hydride ion) can move to an adjacent carbon atom if a more stable carbocation results. The same carbon skeleton remains after a Hydride Shift, but the positive charge is located on a different carbon atom than expected. Alkyl Shift - an alkyl group (even an alkyl portion of a ring) with its pair of bonding electrons can move to an adjacent carbon atom if a more stable carbocation results. A different carbon skeleton results after an Alkyl Shift, and both the alkyl group and the positive charge are located on a different carbon atoms than expected.If an Alkyl Shift occurs in a ring, a larger ring will form ("ring-expansion"). If the Alkyl Shift can be a normal alkyl group or a ring-expansion, show the ring-expansion product. If both the Hydride Shift and the Alkyl Shift are possible, show the one that forms the MOST stable carbocation. Use ChemSketch to draw the rearranged carbocation of the structure below. If a rearrangement is not likely to occur (i.e. the molecule is not structured to take advantage of either of the two shifts mentioned above), enter "None" without the quotes.
none
What alkene would you react with HBr to exclusively form the alkyl bromide shown below?Draw the alkene in ChemSketch and enter its name below. If the alkyl bromide cannot be formed (or formed exclusively) from any alkene using just HBr, enter the word "None" - without the quotes.
none
same as above
none
Use ChemSketch to draw the Least Stable Pentene isomer (or stereoisomer) of C5H10
pent-1-ene
Use ChemSketch to draw the alkene constitutional and stereoisomers of C5H10
pent-1-ene (2E)-pent-2-ene (2Z)-pent-2-ene 2-methylbut-1-ene 3-methylbut-1-ene 2-methylbut-2-ene
Use ChemSketch to generate the names of the C4H8 AND C5H10 alkene isomers that CAN NOT have a cis / trans (or E / Z) stereoisomer.
pent-1-ene 2-methylbut-1-ene 3-methylbut-1-ene 2-methylbut-2-ene 2-methylprop-1-ene but-1-ene