Question
Propose a synthesis of cis-4-octene starting from 1-pentyne. Specify the reagents you would use to carry it out. If two or more routes to the same product are possible. show only on of them. (Spectify reagents using letters from the table, the first reagent on the left. Example: ab) Reagents available a. 1. \( \mathrm{BH}_{3} \) d. \( 1 . \mathrm{Hg}(\mathrm{OAc})_{2}, \mathrm{H}_{2} \mathrm{O} \) 2. \( \mathrm{CH}_{3} \mathrm{COOH} \) 2. \( \mathrm{NaBH}_{4} \) g. \( \mathrm{H}_{2} / \mathrm{Ni} \) b. \( \mathrm{H}_{2} \mathrm{O}, \mathrm{H}_{2} \mathrm{SO}_{4}, \mathrm{HgSO}_{4} \) e. 1. \( \mathrm{NaNH}_{2} / \mathrm{NH}_{3} \) 2. \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} \) h. \( 2 \mathrm{~mol} \mathrm{Na} / \mathrm{NH}_{3}( \) ( ) c \( \mathrm{O}=\mathrm{O}_{4}, \mathrm{H}_{2} \mathrm{O}_{2} \) 1. \( \mathrm{H}_{2} \) / Lindlar catalyst The reagents are:
Ask by Stewart Bryant. in the United States
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
Start with 1-pentyne and use reagent **e** (NaNH₂/NH₃ followed by CH₃CH₂CH₂Br) to form 4-octyne. Then, reduce 4-octyne with H₂/Lindlar catalyst to obtain cis-4-octene.
Solution
1. Start with 1‐pentyne,
\[
\mathrm{HC} \equiv \mathrm{CCH_2CH_2CH_3}
\]
and treat it with reagent **e**. This involves two steps: first deprotonation with
\(
\mathrm{NaNH_2/ NH_3}
\) to form the acetylide anion, and then alkylation with
\(
\mathrm{CH_3CH_2CH_2Br}
\). The overall transformation gives
\[
\mathrm{CH_3CH_2CH_2-C} \equiv \mathrm{C-CH_2CH_2CH_3}
\]
which is 4‐octyne.
2. Reduce 4‐octyne with the Lindlar catalyst by using the reagent
\(
\mathrm{H_2}\ /\ \mathrm{Lindlar\ catalyst}
\)
to selectively hydrogenate the triple bond to a cis double bond. This yields cis‐4‐octene:
\[
\mathrm{CH_3CH_2CH_2-CH=CH-CH_2CH_2CH_3}
\]
Thus, the synthesis proceeds via reagent **e** (for alkylation) followed by \( \mathrm{H_2}\ /\ \mathrm{Lindlar} \) (for partial hydrogenation).
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Simplify this solution Beyond the Answer
To synthesize cis-4-octene from 1-pentyne, you can follow these steps: 1. First, use reagent \( e \) (1. \( \mathrm{NaNH}_{2} / \mathrm{NH}_{3} \); 2. \( \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Br} \)) to perform an alkyne substitution, converting 1-pentyne into 1-octyne. This forms a longer carbon chain by introducing the butyl group. 2. Next, use reagent \( h \) (2 mol \( \mathrm{Na} / \mathrm{NH}_{3} \)) to partially hydrogenate 1-octyne to yield cis-4-octene. This method selectively reduces the triple bond to a cis double bond. So the sequence of reagents you would use is: \( e \) followed by \( h \).
