E1 and E2 Flashcards

1
Q

Rate of E2?

A

halogenoalkane and base

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2
Q

Stereoselectivity of E2?

A

When halogenoalkanes can adopt two antiperiplanar conformation the E isomer is formed in higher yield, more stable because of less steric strain

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3
Q

E2 Saytsev elimination?

A

Aliphatic halogenoalkanes that can rotate freely about C-C bonds normally produce most substituted alkene, the more stable the alkene the more stable the transition state

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4
Q

E2 Hoffmann elimination?

A

Less stable alkene is formed selectively

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5
Q

Rate of E1?

A

Halogenoalkane only

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6
Q

Why does E1 not require anitperiplanar conformation?

A

C-C bond in intermediate carbocation can rotate freely to align sigma orbital of C-H bond with empty p orbital, when orbitals are parallel they can overlap side on and the sp3 carbon atom next to the carbocation changes hybridisation to become an sp2 carbon atom and form the C=C pi bond

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7
Q

Alkyl groups E1?

A

More alkyl groups faster E1, +I groups so stabilise the intermediate carbocation, more stable carbocation the faster it is formed by cleavage of C-X bond hence why tertiary halogenoalkanes readily undergo E1, secondary are slower and primary do no undergo E1

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8
Q

Regioselectivity of E1?

A

If alkene can form two structural isomers the major product is most substituted isomer, since transition state leading to it is more substituted and more stabilised and formed faster

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9
Q

Stereoselectivity of E1?

A

E alkene usually formed, transition state leading to E alkene has lower energy and so formed faster

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10
Q

Base on rate?

A

Increased concentration or increased basicity only increases E2 not E1

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11
Q

Effect of solvent on rate SN1?

A

Increasing solvent polarity increases rate significantly polar solvents solvate the carbocation and halide ions, changing from polar protic to polar aprotic decreases rate since aprotic solvents solvate carbocation but not halide ion

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12
Q

Effect of solvent on rate SN2?

A

Increasing solvent polarity slightly decreases the rate since polar solvents solvate the base, changing from polar protic to polar aprotic increases rate since aprotic solvents do not solvate the base

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13
Q

E2 vs SN2?

A

Tertiary halogenoalkanes when reacting in conditions that favour E2/SN2 reactions will only form elimination products

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14
Q

Increasing temperature SN2/E2?

A

Increases both SN2 and E2, the rate of E2 increases more than SN2 so higher temperature favours formation of elimination products, E2 has higher activation energy so higher temperatures are required

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15
Q

Increasing temperature SN1/E1?

A

Increases both SN1 and E1, the rate of E1 increases more than SN1 so higher temperature favours formation of elimination products, E1 has higher activation energy so higher temperatures are required

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16
Q

Solvents?

A

SN1/E1 products favoured in polar protic solvents

17
Q

Nucleophiles/base?

A

SN2/E2 products favours with high concentration of strong nucleophile or base