Q41 E - Organic Chemistry (Mcmurry)

Question

Question: The syntheses shown here are unlikely to occur as written. What is

wrong with each?

                                                   Reaction 1

                                                      Reaction 2

Step-by-Step Solution

Answer

Reaction 1 is wrong because Reaction 3 fluoroalkanes don’t usually form Grignard reagents.

Reaction 2 is wrong because cycloalkene have two allylic position for bromination

Reaction 3 is wrong because fluoroalkanes don’t react with lithium dialkyl copper.

Step-by-Step Solution Step 1: Alkyl halide reaction with Mg and H 3 O +

Alkyl halides react with magnesium to give Grignard reagents and hydronium ion reaction with Grignard reagents gives alkane. Fluoroalkanes don’t usually form Grignard reagents.

Step 2: Alkene reaction with N B S ,   h ν  and CCl 4

When NBS and heat are used on an alkene it gives allylic bromine. NBS induces allylic bromination. Each cycloalkene has two allylic positions.

Step 3: Alkyl halide reaction with lithium dialkyl copper

Alkyl halide and lithium dialkyl copper reacts to give a higher hydrocarbon chain. Fluoroalkanes don’t react with lithium dialkyl copper.

Step 4: Reaction 1 is wrong because

                                                        Reaction 1

Reaction 1 is wrong because fluoroalkanes don’t usually form Grignard reagents and here fluoroalkane is reacting with magnesium.

Step 5: Reaction 2 is wrong because

                                                          Reaction 2

 

Reaction 2 is wrong because cycloalkenes have two allylic position for bromination and here only one product is given. The two products are:

Two products of allylic bromination

Step 6: Reaction 3 is wrong because

                                                     Reaction 3

Reaction 3 is wrong because fluoroalkanes don’t don’t react with lithium dialkyl copper and here it is shown that fluoroalkane react with lithium dialkyl copper.