Look at the given reaction and the provided compounds. Fill in the blanks with t

Question

Look at the given reaction and the provided compounds. Fill in the blanks with the single letter code of the appropriate product compound/intermediate, or the appropriate descriptive vocabulary word/phrase to complete the description. Remember the computer is very literal so check your spelling carefully!

What is the major product of the reaction?
What are the minor products of the reaction? mostly compound___ and a small amount of compound______ .
This reaction follows _____ regioselectivity and uses a(n) ______ mechanism

If we convert the reagents to (CH3)3COK, (CH3)3COH

What is the major product of the reaction? Compound _____
What is the minor products of the reaction? Compound ______and______
This reaction follows ______ regioselectivity and uses a(n)_______ mechanism

If we convert the Br in the starting material to OH, and the reagent to H2SO4

What are the major products of the reaction assuming no rearrangement? Compounds_____ and_____
What are the minor products of the reaction assuming no rearrangement? Compound_____ and______

If a rearrangement occurs what additional products will form? Compounds______ and_____ , of which compound______ is the most likely due to stability.
This reaction follows a(n)______ mechanism.

Explanation / Answer

I) Reagents CH3CH2ONa, CH3CH2OH

Sodium ethoxide is a strong base, hence it would follows E2 elimination reaction and the Major product will be "B". which is actually a rearranged product of "E".

Minor products will be "A" and a small amount of "J" will also be possible. "J" follows SN2 mechanism.

The more stable product is one which has more substituted double bond (i,e; "B"). And the formation of "B" follows E2 mechanism.

II) Reagents (CH3)3COK, (CH3)3COH

which also a strong base and a bulky reagent as compared with the previous one.

The ultimate product will be "B" only. There may be a small amount of "A" possible.

It won't undergo SN2 reaction due to the bulkiness of the reagent.

Here also the regioselectivity goes to more substituted double bond only. "B" formation through E2 mechanism.

III) Br is changed as OH and the reagent H2SO4

-OH never be a leaving group in E2 reaction. The first step will be the protonation of -OH, which forms oxonium ion. then the adjucent Hydrogen to oxonium ion will abstracted and H2O will leave from the molecule and form a new double dond.

Here too "B" & "A" will be the major product (after rearrangement).

"E" & "F" are the minor products before rearrangement.

After rearrangement "B" & "G" are the major possible products. Between these "G" tetra substituted alkene is a most stable product (Zaitsev's Product).

Unlike previous reactions this follows E1 mechanism. The formation of oxonium intermediate is a slow step then further abstraction of Hydrogen & H2O elimination is a fast step.

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