Dropdown Menus:
1. methyl, primary (unhindered), primary (hindered), secondary, tertiary
2. poor, weakly basic, strongly basic (hindered), strongly basic (unhindered)
3. protic, aprotic
4. SN2, SN1, E1, E2, SN2 AND E2, SN1 AND E1, No Reaction
5. randomized, inverted, predictable but not inverted, unchanged
6. a planar carbocation intermediate, backside attack transition state, anti-periplanar transition state, other
Homework Answers
The base used ( sodium ethoxide in ethanol) is a very good one and it is also a very good neucleophile
So the substrate given will undergo E2 reaction (as the substrate contains beta hydrogen atoms and the base used is good one)
And it will also undergo SN2 reaction( very little) as the neucleophile is a good one and bromide is a good leaving group...
But the SN2 reaction is not as good as E2 as the reaction centre is secondary one and in the neucleophile. ( ethoxide) is so small also ( rate of SN2 reaction with the carbo center- methyl>primary> secondary> tertiary)
Si it will undergo E2 preferably more easily than SN2
So the E2 products will be the major and if SN2 product is obtained that will be in a very very small amount....
Not SN1 or E1 as the base/neucleophile is a very good one but for SN1 or E1 reaction the strong base or neucleophile is not a requirement as it is not involved in the rate determining step of those reactions....
Also the beta Hyde atoms are also in anti configuration with the leaving group bromide which an important requirment for E2 reaction....
Add Answer to:
Dropdown Menus: 1. methyl, primary (unhindered), primary (hindered), secondary, tertiary 2. poor, weakly basic, strong...
Dropdown Menus: 1. methyl, primary (unhindered), primary (hindered), secondary, tertiary 2. poor, weakly basic, strongly basic...
Dropdown Menus:
1. methyl, primary (unhindered), primary (hindered), secondary,
tertiary
2. poor, weakly basic, strongly basic (hindered), strongly basic
(unhindered)
3. protic, aprotic
4. SN2, SN1, E1, E2, SN2 AND E2, SN1 AND E1, No Reaction
5. randomized, inverted, predictable but not inverted,
unchanged
6. a planar carbocation intermediate, backside attack transition
state, anti-periplanar transition state, other
OH Hom 유 용 The Tone The ♡ ♡ Tñ y OH H OH E OH F HO3 G Look at the given...
Look at the given reaction and select the most appropriate answer from the dropdown menus to...
Look at the given reaction and
select the most appropriate answer from the dropdown menus to fill
in the analysis.
The substrate haloalkane is a ["methyl", "primary (unhindered)",
"primary (hindered)", "secondary", "tertiary"] haloalkane. The
nucleophile is a ["poor", "weakly basic", "strongly basic
(unhindered)", "strongly basic (hindered)"] nucleophile. And the
solvent is ["protic", "aprotic"].
This means the reaction likely follows a(n) ["SN2", "SN1", "E2",
"E1", "SN2 and E2", "SN1 and E1", "No reaction"] mechanism(s),
which will result in the stereochemistry at...
Look at the given reaction and select the most appropriate answer from the dropdown menus to...
Look at the given reaction and select the most appropriate
answer from the dropdown menus to fill in the analysis. Choose one
of the answers in the bracket:
The substrate haloalkane is a ["methyl", "primary (unhindered)",
"primary (hindered)", "secondary", "tertiary"] haloalkane. The
nucleophile is a ["poor", "weakly basic", "strongly basic
(unhindered)", "strongly basic (hindered)"] nucleophile. And the
solvent is ["protic", "aprotic"] .
This means the reaction likely follows a(n) ["SN2", "SN1", "E2",
"E1", "SN2 and E2", "SN1 and E1", "No...
For the first six blanks, please choose only options from the dropdown menus provided below the...
For the first six blanks, please choose only options from the
dropdown menus provided below the pictures.
Dropdown Menus:
1. methyl, primary (unhindered), primary (hindered), secondary,
tertiary
2. poor, weakly basic, strongly basic (hindered), strongly basic
(unhindered)
3. protic, aprotic
4. SN2, SN1, E1, E2, SN2 AND E2, SN1 AND E1, No Reaction
5. randomized, inverted, predictable but not inverted,
unchanged
6. a planar carbocation intermediate, backside attack transition
state, anti-periplanar transition state, other
CH3OH ~ ??? Br - în...
For the first six blanks, please choose options only from the dropdown menus provided below the...
For the first six blanks, please choose options only from the
dropdown menus provided below the pictures.
Dropdown Menus:
1. methyl, primary (unhindered), primary (hindered), secondary,
tertiary
2. poor, weakly basic, strongly basic (hindered), strongly basic
(unhindered)
3. protic, aprotic
4. SN2, SN1, E1, E2, SN2 AND E2, SN1 AND E1, No Reaction
5. randomized, inverted, predictable but not inverted,
unchanged
6. a planar carbocation intermediate, backside attack transition
state, anti-periplanar transition state, other
OMS Low Caucomm tom bra tron...
Are my answers right??? gr mom ??? Su Tone ho a Toth Look at the given reaction and select the most appropriate...
Are my answers right???
gr mom ??? Su Tone ho a Toth Look at the given reaction and select the most appropriate answer from the dropdown menus to fill in the analysis. The substrate haloalkane is a secondary haloalkane. The Look at the given reaction and select the most appropriate answer from the dropdown menus to fill in the analysis. The substrate haloalkane is a secondary haloalkane. The nucleophile is a poor nucleophile. And the solvent is protic This means...
Organic Chemistry: Vocabulary: i.) If we have a weakly basic nucleophile and a secondary haloalkane we’ll...
Organic Chemistry: Vocabulary: i.) If we have a weakly basic nucleophile and a secondary haloalkane we’ll generally get _ mechanism(s). a. SN1 only b. SN2 c. E2 d. E1 only e. SN1 and E1 f. no reaction This mechanism will have ____ stereochemistry, a. predictable b. random due to the ____. a. planar intermediate b. backside attack transition state c. frontside attack transition state d. anti-periplanar transition state _____________________________________________________ ii.) Carbocation rearrangements will tend to form the __ stable intermediate....
Organic Chemistry: Vocabulary I.) In general elimination reactions will form the _ stable alkene. a. more...
Organic Chemistry: Vocabulary I.) In general elimination reactions will form the _ stable alkene. a. more b. less This is typically the alkene with: a. more substituents b. less substituents. c. cannot tell The exception is when we have an __ mechanism with a _. a. E1, hindered poor nucleophile b. E1, unhindered poor nucleophile c. E2, hindered strongly basic nucleophile d. E2, unhindered strongly basic nucleophile ________________________________________________________________________________________________ II.) Staggered rotamers have a _________ degree torsional angle. Nucleophiles are ___...
2. Predict the product(s) of each of the reaction conditions below. Specify the substrate type (primary,...
2. Predict the product(s) of each of the reaction conditions
below. Specify the substrate type (primary, secondary, or tertiary
alkyl halide), nucleophile (strong or poor), and the solvent type
(protic or aptrotic). Specify the type of reaction mechanism (SN1
or SN2), show the curved arrow mechanism and draw the products as
3D structures to illustrate stereochemistry.
Predict the product(s) of each of the reaction conditions below. Specify the substrate type (1º, 2º, or 3 alkyl halide). mucleophile (strong or poor),...
1. In both the sodium iodide test and the silver nitrate test, why does 2-bromobutane react faster...
1. In both the sodium iodide
test and the silver nitrate test, why does 2-bromobutane react
faster than 2-chlorobutane?
Bromine is a better leaving group since it is a weaker base than
chlorine is.
2. a. Why does benzyl chloride react under both
SN1 and SN2 conditions?
Benzyl
chloride is a primary alkyl halide, hence reactive under SN2
conditions.
The
primary carbocation formed due to the departure of Cl- is
stabilized by the pi electrons in the benzene ring.
b. Why is...
Dropdown Menus:
1. methyl, primary (unhindered), primary (hindered), secondary,
tertiary
2. poor, weakly basic, strongly basic (hindered), strongly basic
(unhindered)
3. protic, aprotic
4. SN2, SN1, E1, E2, SN2 AND E2, SN1 AND E1, No Reaction
5. randomized, inverted, predictable but not inverted,
unchanged
6. a planar carbocation intermediate, backside attack transition
state, anti-periplanar transition state, other
OH Hom 유 용 The Tone The ♡ ♡ Tñ y OH H OH E OH F HO3 G Look at the given...
Look at the given reaction and
select the most appropriate answer from the dropdown menus to fill
in the analysis.
The substrate haloalkane is a ["methyl", "primary (unhindered)",
"primary (hindered)", "secondary", "tertiary"] haloalkane. The
nucleophile is a ["poor", "weakly basic", "strongly basic
(unhindered)", "strongly basic (hindered)"] nucleophile. And the
solvent is ["protic", "aprotic"].
This means the reaction likely follows a(n) ["SN2", "SN1", "E2",
"E1", "SN2 and E2", "SN1 and E1", "No reaction"] mechanism(s),
which will result in the stereochemistry at...
Look at the given reaction and select the most appropriate
answer from the dropdown menus to fill in the analysis. Choose one
of the answers in the bracket:
The substrate haloalkane is a ["methyl", "primary (unhindered)",
"primary (hindered)", "secondary", "tertiary"] haloalkane. The
nucleophile is a ["poor", "weakly basic", "strongly basic
(unhindered)", "strongly basic (hindered)"] nucleophile. And the
solvent is ["protic", "aprotic"] .
This means the reaction likely follows a(n) ["SN2", "SN1", "E2",
"E1", "SN2 and E2", "SN1 and E1", "No...
For the first six blanks, please choose only options from the
dropdown menus provided below the pictures.
Dropdown Menus:
1. methyl, primary (unhindered), primary (hindered), secondary,
tertiary
2. poor, weakly basic, strongly basic (hindered), strongly basic
(unhindered)
3. protic, aprotic
4. SN2, SN1, E1, E2, SN2 AND E2, SN1 AND E1, No Reaction
5. randomized, inverted, predictable but not inverted,
unchanged
6. a planar carbocation intermediate, backside attack transition
state, anti-periplanar transition state, other
CH3OH ~ ??? Br - în...
For the first six blanks, please choose options only from the
dropdown menus provided below the pictures.
Dropdown Menus:
1. methyl, primary (unhindered), primary (hindered), secondary,
tertiary
2. poor, weakly basic, strongly basic (hindered), strongly basic
(unhindered)
3. protic, aprotic
4. SN2, SN1, E1, E2, SN2 AND E2, SN1 AND E1, No Reaction
5. randomized, inverted, predictable but not inverted,
unchanged
6. a planar carbocation intermediate, backside attack transition
state, anti-periplanar transition state, other
OMS Low Caucomm tom bra tron...
Are my answers right???
gr mom ??? Su Tone ho a Toth Look at the given reaction and select the most appropriate answer from the dropdown menus to fill in the analysis. The substrate haloalkane is a secondary haloalkane. The Look at the given reaction and select the most appropriate answer from the dropdown menus to fill in the analysis. The substrate haloalkane is a secondary haloalkane. The nucleophile is a poor nucleophile. And the solvent is protic This means...
2. Predict the product(s) of each of the reaction conditions
below. Specify the substrate type (primary, secondary, or tertiary
alkyl halide), nucleophile (strong or poor), and the solvent type
(protic or aptrotic). Specify the type of reaction mechanism (SN1
or SN2), show the curved arrow mechanism and draw the products as
3D structures to illustrate stereochemistry.
Predict the product(s) of each of the reaction conditions below. Specify the substrate type (1º, 2º, or 3 alkyl halide). mucleophile (strong or poor),...
1. In both the sodium iodide
test and the silver nitrate test, why does 2-bromobutane react
faster than 2-chlorobutane?
Bromine is a better leaving group since it is a weaker base than
chlorine is.
2. a. Why does benzyl chloride react under both
SN1 and SN2 conditions?
Benzyl
chloride is a primary alkyl halide, hence reactive under SN2
conditions.
The
primary carbocation formed due to the departure of Cl- is
stabilized by the pi electrons in the benzene ring.
b. Why is...
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