The synthesis of 3-octyne is achieved by addi | vedclass

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(D) The synthesis of $3-$octyne $(CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3)$ involves the alkylation of an acetylide ion. First,$1-$butyne $(CH_3CH_2C \equi

Vedclass · Accepted Answer

$BrCH_2CH_2CH_2CH_3$ and $CH_3CH_2C \equiv CH$

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(D) The synthesis of $3-$octyne $(CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3)$ involves the alkylation of an acetylide ion. First,$1-$butyne $(CH_3CH_2C \equiv CH)$ reacts with sodium amide $(NaNH_2)$ to form the sodium butynide nucleophile: $CH_3CH_2C \equiv CH + NaNH_2 \rightarrow CH_3CH_2C \equiv C^{-}Na^{+} + NH_3$. Next,this nucleophile performs an $S_N2$ reaction with $1-$bromobutane $(CH_3CH_2CH_2CH_2Br)$: $CH_3CH_2C \equiv C^{-}Na^{+} + CH_3CH_2CH_2CH_2Br \rightarrow CH_3CH_2C \equiv CCH_2CH_2CH_2CH_3 + NaBr$. Thus,the required reagents are $1-$bromobutane and $1-$butyne.

The synthesis of $3-$octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne. The bromoalkane and alkyne respectively are

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Which one of the following statements is false?

In the given reaction $x$ is:
$CH_3 - C \equiv CH$ $\xrightarrow{Excess\ HCl} (A)$ $\xrightarrow{H_2O/OH^{-}} (x)$

The product $(K)$ of the following reaction is:

In the reaction
$HC \equiv CH \xrightarrow[(ii) CH_3CH_2Br]{(i) NaNH_2 / liq. NH_3} X$
$\xrightarrow[(ii) CH_3CH_2Br]{(i) NaNH_2 / liq. NH_3} Y$
$X$ and $Y$ are

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