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4. (31 pts) Provide the reaction products and indicate the type of reactions ( I, UL,...

Question:

4. (31 pts) Provide the reaction products and indicate the type of reactions ( I, UL, TE EZ) and optical activity (Yes or No)
4. (31 pts) Provide the reaction products and indicate the type of reactions ( I, UL, TE EZ) and optical activity (Yes or No) in the round rectangles, respectively. If reaction does not take place, indicate "No Reaction" in the box Draw 2 major products Reaction type Snd Draw 1 minor product Reaction type EL VW Is the product optically active? ha Draw 1 product Reaction type EZ Сн,он NaCN acetone Is the product optically active? ho Br Draw 1 product Reaction type secondary alkyl-bromide LDA isopropyl amine NaOCOCHE DMF 20. CH. Draw 1 product Reaction type E2 NaOCH CH, OH Draw 1 MINOR product Draw 1 MAJOR product Reaction type EZ Reaction type SNZ OCH3

Answers

Reaction Mechanism:

SN1 reaction: If tertiary carbocation formed after the leaving group leaving or alkyl migration with weak nucleophile then the reaction proceeds via an SN1 mechanism. SN1 reaction will give the racemic mixture of product.

SN2 reaction: If the nucleophile is strong and the reactant has alkyl-substitution on primary or secondary carbon then reaction proceeds via an SN2 mechanism. SN2 mechanism gives an inversed( If you take R- isomer in the reactant you will get S- isomer in the product) optically active product.

E2 reaction: If the base was very strong and the reactant has alkyl-substitution on primary or secondary carbon with alpha-hydrogen then reaction proceeds via an E2 reaction.

CH₂OH 2° -Carbocation 3- carbocation KY CH2O OCH2 Осн: OCH3 SN2 Mimor product) cptically active compound (Major products) racna OCS CH₃OH alkone (Major optically inactive) product) Baoch X ochz CH₂OH optically active (minnor procluet)

In the first reaction, methanol is a weak nucleophile so SN2 reaction, not a preferable reaction. So SN2 reaction will give the minor product. And the SN1 reaction will give the major product.

In the second reaction cyanide anion formed from NaCN act as a good nucleophile for an SN2 reaction.

So the reaction will give the optically active cyano substituted product.

In the third reaction, LDA is a strong base which will give only E2 pathway product (alkene)

In the fourth reaction, acetate anion from (NaOCOCH3) is moderate nucleophile so it will give SN2 pathway product.

In the fifth reaction, sodium methoxide is a strong base which will give the E2 reaction product (alkene) as a major product. and methoxide anion gives some amount of SN2 pathway product also.

  

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