Question

SN square reaction will be fastest in

Answer 1

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( for mechanism of sn2 see attached file 1)

♦♦♦• Reaction is:

Stereospecific (Walden Inversion of configuration)

Concerted - all bonds form and break at same time

Bimolecular - rate depends on concentration of both nucleophile and substrate

♦♦♦• Substrate:

Best if primary (one substituent on carbon bearing leaving group)

works if secondary, fails if tertiary

♦♦♦• Nucleophile:

Best if more reactive (i.e. more anionic or more basic)

♦♦♦• Leaving Group: Best if more stable (i.e. can support negative charge well):

TsO- (very good) > I- > Br- > Cl- > F- (poor)

RF , ROH , ROR , RNH2


♦♦♦• Solvent:

Polar Aprotic (i.e. no OH) is best.

For example dimethylsulfoxide ( CH3 ( HCON(CH3)2 ), acetonitrile ( CH3

Protic solvents (e.g. H2 but can be used in some case are NEVER Substrates for SN2 reactions

Leaving Groups on double-bonded carbons are never replaced by SN2 reactions SOCH3 ), dimethylformamide CN ). O or ROH) deactivate nucleophile by hydrogen bonding


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( for mechanism of sn1 see attached file 2)

♦♦♦• Reaction is:

Non-stereospecific (attack by nucleophile occurs from both sides)

Non-concerted - has carbocation intermediate

Unimolecular - rate depends on concentration of only the substrate

♦♦♦• Substrate:

Best if tertiary or conjugated (benzylic or allylic) carbocation can be formed as leaving group departs

never primary


♦♦♦• Nucleophile:

Best if more reactive (i.e. more anionic or more basic)

♦♦♦• Leaving Group: Same as SN2

best if more stable (i.e. can support negative charge well)

Examples: TsO- (very good) > I- > Br- > Cl- > F- (poor)

However, tertiary or allylic ROH or ROR' can be reactive under strongly acidic conditions to replace OH or OR

♦♦♦• Solvent:

Same as SN2

Polar Aprotic (i.e. no OH) is best

Examples: dimethylsulfoxide ( CH3 ( HCON(CH3)2 ), acetonitrile ( CH3

Protic solvents (e.g. H2
SOCH3 ), dimethylformamide CN ).
O or ROH) deactivate but can be used in some cases

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