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Question: Predict the product of the reaction. ```latex \begin{array}{ccc} & CH_3 & H\\ & | & |\\ CH_3 - C & ...

Predict the product of the reaction.

\begin{array}{ccc}
& CH_3 & H\\
& | & |\\
CH_3 - C & - & C - CH_2 - CH = CH_2\\
& | & |\\
& H & Cl
\end{array}
\xrightarrow{alc.KOH} ?
A

(CH3)2C=CH−CH2−CH=CH2(CH_3)_2C=CH-CH_2-CH=CH_2

B
\begin{array}{c}
(CH_3)_2CH - CH = C - CH = CH_2\\
|\\
CH_3
\end{array}
C

(CH3)2CH−CH=CH−CH=CH2(CH_3)_2CH - CH = CH - CH = CH_2

D
\begin{array}{c}
(CH_3)_2C = CH - C = CH_2\\
|\\
CH_3
\end{array}
Answer

(CH3)2C=CH−CH2−CH=CH2(CH_3)_2C=CH-CH_2-CH=CH_2

Explanation

Solution

We start with the compound

CH3-C(CH3)(H)-C(H)(Cl)-CH2-CH=CH2\text{CH}_3\text{-C(CH}_3\text{)(H)-C(H)(Cl)-CH}_2\text{-CH=CH}_2

In the presence of alcoholic KOH, an E2 elimination takes place. The base abstracts a β‐hydrogen which is anti-periplanar to the chlorine. There are two possible β‑positions: one from the carbon to the left (C₂) and one from the right (C₄). Since the hydrogen on C₂ is on the more highly substituted carbon (C₂ is bonded to two methyl groups), removal of this hydrogen yields the more substituted (Zaitsev) alkene.

Thus, the elimination of H from C₂ and Cl from C₃ gives the double bond between C₂ and C₃, producing

(CH3)2C=CH−CH2−CH=CH2(CH_3)_2C=CH-CH_2-CH=CH_2