$\ce{4-phenyl~butene~ring~extension}$

Chemistry Level 3

Starting with $\ce{4-phenyl~butene}$ , you proceed to react it in the following order:

$\ce{HCl}$
$\ce{AlCl}_3$
$\ce{Br}_2$ in $\ce{light}$
$\ce{Lithium~Di-isopropylaminde}$

Among choices $A$ through $D$ , choose the major product of the above series of reactions.

A

B

C

D

1 solution

David Hontz
Jul 14, 2016

$HCl$ is added to the double bond, with $Cl$ attaching to the most substituted carbon in following Markovnikov's Rule
$AlCl_3$ causes Friedel Craft Alkylation to occur, causing the chlorine-bonded carbon to form a ring onto benzene
$Br_2$ will then add to the preferred $3^{\circ}$ carbon. Note, $light$ is not energetic enough to add $Br$ to an aromatic group like benzene
$LDA$ will remove $Br$ to form a terminal double bond that is in conjugation with $benzene$ . The least substituted double bond is preferred because $LDA$ is a bulky reagent

$\ce{Major~Product} \rightarrow \boxed{B}$

well,the tricky part is to understand that a 6C ring would not form.

A Former Brilliant Member - 3 years, 11 months ago

Your right. It should be noted that a $2^{\circ}$ carbocation would form which is more stable than the $1^{\circ}$ carbocation that would be needed to form a six-member ring.

David Hontz - 3 years, 10 months ago

Oh i forgot that LDA was a bulky base!

Nice question!

Harsh Shrivastava - 3 years, 11 months ago

4 - p h e n y l b u t e n e r i n g e x t e n s i o n \ce{4-phenyl~butene~ring~extension} 4 - p h e n y l b u t e n e r i n g e x t e n s i o n

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$\ce{4-phenyl~butene~ring~extension}$