I don\'t need to answers to the problems. I just don\'t understand part 2. How i
ID: 1058939 • Letter: I
Question
I don't need to answers to the problems. I just don't understand part 2. How is it determined what reactions will produce both elimination and substitution products? Please explain.
1. Predict the elimination products of the following reactions. When two alkenes are possi- ble, predict which one will be the major product. Explain your answers, showing the degree of substitution of each double bond in the products. A2. Which of these reactions are likely to produce both elimination and substitution products? (a) 2-bromopentane NaOCH (b) 3-bromo-3-methylpentane NaOMe (Me methyl, CH3) (c) 2-bromo-3-ethylpentane NaOH (d) cis-1-bromo-2-methylcyclohexane NaOEt (Et ethyl, CH2CH3Explanation / Answer
The decesion for substitution vs elimination depends upon many factors such as alkyl halide , type of base/nucleophile, temperature and solvent used for reaction. Follow these guidelinesas:
The SN1 reaction compete to E1 and SN2 to E2.
SN1 reaction:
SN1 reaction is an unimolecular and hence a first order reaction. So, only substrate such as electrophile affects the reaction rate.
For reaction: RX + Y ---> RY + X
The reaction rate is expressed as rate = k [RX]
This reaction has two steps.
Intermediate is a stable carbocation which is formed during the course of reaction.
This generally has two transition states as it is a two step reaction.
The stability of carbocation is the key factor of the reaction. Hence, reactivity of alkyl halides are as: 3o > 2o > 1o
For this reaction, weak or neutral nucleophiles are sufficient.
Favoured by, polar protic solvents as they can stabilize the carbocation intermediates.
Stereochemical outcome is a racemic mixture of product.
SN2 reaction:
It is a bimolecular or a second order reaction. Hence, both substrate such as electrophile and nucleophile affect the rate of reaction
For reaction: RX + Y ---> RY + X
The reaction rate is expressed as rate = k[RX][Y]
This reaction has only one step.
This reaction goes through a transition state, there is no intermediate cation formation.
Presence of steric hindrance affect the reaction as transition state becomes more crowded. Hence reactivity of alkyl halides are as: 1o > 2o > 3o
In this pathway strong nucleophiles are desired.
The reaction is favoured in polar aprotic solvents eg. DMF, DMSO and acetone.
The inversion of configuration is observed (called as Walden inversion) in products.
E1 reaction:
E1 reaction is an unimolecular and hence a first order reaction. So, only substrate such as electrophile affects the reaction rate not the nucleophile.
For reaction: RX + Y ---> RY + X
The reaction rate is expressed as rate = k [RX]
This reaction has two steps.
Intermediate is a stable carbocation which is formed during the course of reaction.
This generally has two transition states as it is a two step reaction.
The stability of carbocation is the key factor of the reaction. Hence, reactivity of alkyl halides are as: 3o > 2o > 1o
For this reaction, weak or neutral bases are sufficient.
Favoured by, polar protic solvents as they can stabilize the carbocation intermediates.
Favoured at higher temperature.
Stereochemical outcome is a racemic mixture of product.
E2 reaction:
It is a bimolecular or a second order reaction.
The reaction rate is expressed as rate = k[RX][Y]
This reaction has only one step.
This reaction goes through a transition state; there is no intermediate cation formation.
Presence of steric hindrance affect the reaction as transition state becomes more crowded. Hence reactivity of alkyl halides are as: 1o > 2o > 3o
In this pathway strong bases are desired.
The reaction is favoured in polar aprotic solvents eg. DMF, DMSO and acetone.
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